(a) [tex]4.29\cdot 10^{14}Hz[/tex]
The frequency of an electromagnetic wave is given by
[tex]f=\frac{c}{\lambda}[/tex]
where
c is the speed of light
[tex]\lambda[/tex] is the wavelength
We notice from the formula that the frequency is inversely proportional to the wavelength, so the minimum frequency corresponds to the maximum wavelength, and viceversa
The maximum value of the wavelength of the visible light waves is
[tex]\lambda_{max} = 700 nm = 7.0\cdot 10^{-7} m[/tex] (red light)
so the minimum frequency of visible light is
[tex]f_{min} = \frac{c}{\lambda_{max}}=\frac{3.0\cdot 10^8 m/s}{7.00\cdot 10^{-7}m}=4.29\cdot 10^{14}Hz[/tex]
(b) [tex]7.50\cdot 10^{14}Hz[/tex]
The maximum frequency corresponds to the minimum wavelength;
The minimum wavelength is
[tex]\lambda_{min} = 400 nm = 4.0\cdot 10^{-7} m[/tex] (violet light)
so the maximum frequency of visible light is
[tex]f_{max} = \frac{c}{\lambda_{min}}=\frac{3.0\cdot 10^8 m/s}{4.00\cdot 10^{-7}m}=7.50\cdot 10^{14}Hz[/tex]
(c) 1 m
The wavelength of an electromagnetic wave is given by
[tex]\lambda=\frac{c}{f}[/tex]
as before, we notice that the minimum wavelength corresponds to the maximum frequency, and viceversa.
The maximum frequency of shortwave radio waves is
[tex]f_{max}=300 MHz = 3.0\cdot 10^8 Hz[/tex]
so the minimum wavelength of these waves is
[tex]\lambda_{min} = \frac{c}{f_{max}}=\frac{3.0\cdot 10^8 m/s}{3.0\cdot 10^8 Hz}=1 m[/tex]
(d) 200 m
The minimum frequency of shortwave radio waves is
[tex]f_{min}=1.5 MHz = 1.5\cdot 10^6 Hz[/tex]
so the maximum wavelength of these waves is
[tex]\lambda_{max} = \frac{c}{f_{min}}=\frac{3.0\cdot 10^8 m/s}{1.5\cdot 10^6 Hz}=200 m[/tex]
(e) [tex]6.0\cdot 10^{16}Hz[/tex]
As in part (a) and (b), we can find the frequency of the X-rays by using the formula
[tex]f=\frac{c}{\lambda}[/tex]
The maximum wavelength of the x-rays is
[tex]\lambda_{max} = 5.0 nm = 5.0\cdot 10^{-9} m[/tex]
so the minimum frequency is
[tex]f_{min} = \frac{c}{\lambda_{max}}=\frac{3.0\cdot 10^8 m/s}{5.0\cdot 10^{-9}m}=6.0\cdot 10^{16}Hz[/tex]
(f) [tex]3.0\cdot 10^{19}Hz[/tex]
The maximum frequency corresponds to the minimum wavelength;
The minimum wavelength is
[tex]\lambda_{min} = 1.0\cdot 10^{-2} nm = 1.0\cdot 10^{-11} m[/tex]
so the maximum frequency of the x-rays is
[tex]f_{max} = \frac{c}{\lambda_{min}}=\frac{3.0\cdot 10^8 m/s}{1.0\cdot 10^{-11}m}=3.0\cdot 10^{19}Hz[/tex]
Final answer:
The frequency of electromagnetic waves varies inversely with the wavelength. For visible light, the frequencies range between 4.29 x 10⁹⁴ Hz and 7.5 x 10⁹⁴ Hz; for shortwave radio, the wavelengths range from 1 to 200 meters; for x-rays, the frequencies vary between 6 x 10⁹⁶ Hz and 3 x 10⁹⁹ Hz.
Explanation:
Calculating Frequency and Wavelength of Electromagnetic Waves
To calculate the frequency (ν) and wavelength (λ) of electromagnetic waves, we use the relation c = λν, where c is the speed of light in a vacuum (approximately 3.0 x 10⁸ m/s).
(a) The minimum frequency of visible light (700 nm) is obtained by c/λ, yielding approximately 4.29 x 10¹⁴ Hz.
(b) The maximum frequency for visible light (400 nm) is reached with c/λ, giving around 7.5 x 10¹⁴ Hz.
(c) The minimum wavelength for shortwave radio at 300 MHz is found by c/ν, resulting in 1 meter (m).
(d) The maximum wavelength at 1.5 MHz of shortwave radio is calculated with c/ν, equating to 200 meters (m).
(e) The minimum frequency of x-rays (1.0 x 10⁻² nm) is computed using c/λ, and is approximately 3 x 10¹⁹ Hz.
(f) The maximum frequency for x-rays (5.0 nm) is also obtained with c/λ, yielding around 6 x 10¹⁶ Hz.
When considering gravity acceleration and the force of acceleration, what must be true?
A. The direction of the force and the direction of acceleration must be the same as each other.
B. The mass of the body must be the same as the acceleration of the body.
C. The direction of the force and the direction of acceleration must be opposite of each other.
D. The direction of acceleration must be perpendicular to the direction of the force.
Answer:
A. The direction of the force and the direction of acceleration must be the same as each other.
Explanation:
Force can be defined as push or pull. An unbalanced force that is non-zero net force causes a body to accelerate. Newton's second law states that acceleration depends on the force.
F = m a
where m is the mass of the body and a is the acceleration.
Increase in force causes increase in acceleration. The direction of acceleration and direction of force are same.
Considering acceleration due to gravity and force of acceleration - gravitational force always acts along the line joining the centers of two bodies and so, the direction of the acceleration due to gravity also is in the same direction.
Answer:
The direction of the force and the direction of acceleration must be the same as each other.
Explanation:
According to Newton's second law of motion, the force acting on an object depends on its acceleration. Its formula is given by :
F = m a
Where,
m is the mas of an object
a is its acceleration.
i.e. force is directly proportional to the acceleration of an object. The direction of force and acceleration must be same.
Hence, the correct option is (A) "The direction of the force and the direction of acceleration must be the same as each other".
A 10.0-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.350ohms . Pulling the wire at a steady speed of 4.0m/s causes 4.20W of power to be dissipated in the circuit. part A: How big is the pulling force? part B: What is the strength of the magnetic field?
A) 1.05 N
The power dissipated in the circuit can be written as the product between the pulling force and the speed of the wire:
[tex]P=Fv[/tex]
where
P = 4.20 W is the power
F is the magnitude of the pulling force
v = 4.0 m/s is the speed of the wire
Solving the equation for F, we find
[tex]F=\frac{P}{v}=\frac{4.20 W}{4.0 m/s}=1.05 N[/tex]
B) 3.03 T
The electromotive force induced in the circuit is:
[tex]\epsilon=BvL[/tex] (1)
where
B is the strength of the magnetic field
v = 4.0 m/s is the speed of the wire
L = 10.0 cm = 0.10 m is the length of the wire
We also know that the power dissipated is
[tex]P=\frac{\epsilon^2}{R}[/tex] (2)
where
[tex]R=0.350 \Omega[/tex] is the resistance of the wire
Subsituting (1) into (2), we get
[tex]P=\frac{B^2 v^2 L^2}{R}[/tex]
And solving it for B, we find the strength of the magnetic field:
[tex]B=\frac{\sqrt{PR}}{vL}=\frac{\sqrt{(4.20 W)(0.350 \Omega)}}{(4.0 m/s)(0.10 m)}=3.03 T[/tex]
The strength of the Earth’s magnetic field B at the equator is approximately equal to 5 × 10−5 T. The force on a charge q moving in a direction perpendicular to a magnetic field is given by F = q v B, where v is the speed of the particle. The direction of the force is given by the right-hand rule. Suppose you rub a balloon in your hair and your head acquires a static charge of 9 × 10−9 C. If you are at the equator and driving west at a speed of 90 m/s, what is the strength of the magnetic force on your head due to the Earth’s magnetic field? Answer in units of N. What is the direction of that magnetic force?
The strength of the magnetic force on your head while driving west at the equator is 4.05 x 10^{-5} N, and the direction of this force is upwards, as determined by the right-hand rule.
The force on a charge moving in a magnetic field can be calculated using the formula F = qvB, where F is the force in newtons (N), q is the charge in coulombs (C), v is the velocity in meters per second (m/s), and B is the magnetic field strength in teslas (T). Given the charge on your head is 9 x 10^{-9} C, the Earth's magnetic field strength at the equator is 5 x 10^{-5} T, and your velocity driving west is 90 m/s, the magnitude of the magnetic force can be calculated as follows: F = 9 imes 10^{-9} C x 90 m/s x 5 x 10^{-5} T = 4.05 x 10^{-5} N.
Using the right-hand rule to determine the direction of the force: if you point your thumb in the direction of the velocity (west), and your fingers in the direction of the Earth's magnetic field (north), the force (perpendicular to the palm) will point upwards. Therefore, the direction of the magnetic force on your head is upwards.
When first discovered α, β and γ radioactivity were referred to as α−rays, β-rays and γ- rays. Now we refer to α-particles and β-particles, but still use the term γ-rays. What type of particles are α and β commonly known as? What type of ray is γ commonly known as? What are the signs of the charges on the three forms of radiation?
1. Helium nucleus and electron/positron
- An [tex]\alpha[/tex] decay is a decay in which an [tex]\alpha[/tex] particle is produced.
An [tex]\alpha[/tex] particle consists of 2 protons and 2 neutrons: therefore, in an alpha-decay, the nucleus loses 2 units of atomic number (number of protons) and 2 units of mass number (sum of protons+neutrons).
The alpha-particle consists of 2 protons of 2 neutrons: so it corresponds to a nucleus of helium, which consists exactly of 2 protons and 2 neutrons.
- There are two types of [tex]\beta[/tex] decay:
-- In the [tex]\beta^-[/tex] decay, a neutron decays into a proton emitting a fast-moving electron and an anti-neutrino:
[tex]n \rightarrow p + e^- + \bar{\nu}[/tex]
and the [tex]\beta[/tex] particle in this case is the electron
-- In the [tex]\beta^+[/tex] decay, a proton decays into a neutron, emitting a fast-moving positron and a neutrino:
[tex]p \rightarrow n + e^+ + \nu[/tex]
and the [tex]\beta[/tex] particle in this case is the positron.
2) Gamma ray
A [tex]\gamma[/tex] decay occurs when an unstable (excited state) nucleus decays into a more stable state. In this case, there are no changes in the structure of the nucleus, but energy is released in the form of a photon:
[tex]X^* \rightarrow X + \gamma[/tex]
where the wavelength of this photon usually falls in the part of the electromagnetic spectrum corresponding to the gamma ray region.
So, the [tex]\gamma[/tex] ray is commonly known as gamma radiation.
3)
The sign of the three forms of radiation are the following:
- [tex]\alpha[/tex] particle: it consists of 2 protons (each of them carrying a positive charge of +e) and 2 neutrons (uncharged), so the total charge is
Q = +e +e = +2e
- [tex]\beta[/tex] particle: in case of [tex]\beta^-[/tex] radiation, the particle is an electron, so it carries a charge of
Q = -e
in case of [tex]\beta^+[/tex] radiation, the particle is a positron, so it carries a charge of
Q = +e
- [tex]\gamma[/tex] radiation: the [tex]\gamma[/tex] radiation consists of a photon, and the photon has no charge, so the charge in this case is
Q = 0
What is the equation used to find speed?
Speed=distance/time
Speed=time/distance
Speed=work/distance
Speed=force/time
Answer:
speed =distance/time
Formula
s = d/t
s = speed
d = distance traveled
t = time elapsed
Thanks, Hope this helps.
A ball with a mass of 170 g which contains 4.20×108 excess electrons is dropped into a vertical shaft with a height of 110 m . At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has a magnitude of 0.250 T and direction from east to west. Part A If air resistance is negligibly small, find the magnitude of the force that this magnetic field exerts on the ball just as it enters the field. Use 1.602×10−19 C for the magnitude of the charge on an electron. F F = nothing N Request Answer Part B Find the direction of the force that this magnetic field exerts on the ball just as it enters the field. Find the direction of the force that this magnetic field exerts on the ball just as it enters the field. from north to south from south to north
A) [tex]7.8\cdot 10^{-10} N[/tex]
First of all, we need to find the velocity of the ball as it enters the magnetic field region.
Since the ball starts from a height of h=110 m and has a vertical acceleration of g=9.8 m/s^2 (acceleration due to gravity), the final velocity can be found using the equation
[tex]v^2 = u^2 +2gh[/tex]
where u=0 is the initial velocity. Solving for v,
[tex]v=\sqrt{2(9.8 m/s^2)(110 m)}=46.4 m/s[/tex]
The ball contains
[tex]N=4.20\cdot 10^8[/tex] excess electrons, each of them carrying a charge of magnitude [tex]e=1.602\cdot 10^{-19}C[/tex], so the magnitude of the net charge of the ball is
[tex]Q=Ne=(4.20\cdot 10^8)(1.602\cdot 10^{-19}C)=6.72\cdot 10^{-11}C[/tex]
And given the magnetic field of strength B=0.250 T, we can now find the magnitude of the magnetic force acting on the ball:
[tex]F=qvB=(6.72\cdot 10^{-11}C)(46.4 m/s)(0.250 T)=7.8\cdot 10^{-10} N[/tex]
B) From north to south
The direction of the magnetic force can be found by using the right-hand rule:
- index finger: direction of motion of the ball -> downward
- middle finger: direction of magnetic field --> from east to west
- thumb: direction of the force --> from south to north
However, this is valid for a positive charge: in this problem, the ball is negatively charged (it is made of an excess of electrons), so the direction of the force is reversed: from north to south.
The magnitude of the force that the magnetic field exerts on the ball is 7.812 x 10⁻¹⁰ N.
The magnetic force will directed north to south (reverse direction due to charge of ball).
The given parameters;
mass of the ball, m = 170 g = 0.17 kgexcess electron, N = 4.2 x 10⁸ electronsvertical height of the shaft, h = 110 mmagnetic field strength, B = 0.25 TThe speed of the charge is calculated by applying the principle of conservation of mechanical energy;
Potential energy at top = kinetic energy at bottom
mgh = ¹/₂mv²
2gh = v²
[tex]v = \sqrt{2gh} \\\\v= \sqrt{2\times 9.8 \times 110} \\\\v = 46.43 \ m/s[/tex]
The charge of the excess electron is calculated as follows;
Q = Ne
[tex]Q = (4.2 \times 10^8)\times (1.602 \times 10^{-19})\\\\Q = 6.73 \times 10^{-11} \ C[/tex]
The magnitude of the force that the magnetic field exerts on the ball is calculated as follows;
[tex]F = qvB\\\\F = 6.73 \times 10^{-11} \times 46.43 \times 0.25\\\\F = 7.812 \times 10^{-10} \ N[/tex]
The direction of the magnetic force will be perpendicular to speed of the charge and magnetic field.
The magnetic field is directed to east, the magnetic force will directed south to north. Since the ball is negatively charged (excess electron), the direction will be reversed. Thus, the force will be directed from north to south.
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Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012 W) pulses of electromagnetic waves that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk 4.0 μm in diameter, with the pulse lasting for 5.0 ns with an average power of 1.59×1012 W . We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse.a. How much energy is given to the cell during this pulse? U=___Jb. What is the intensity (in W/m^2) delivered tothe cell? I=___W/m^2c. What is the maximum value of the electric field in the pulse?Emax=___V/md. What is the maximum value of the magnetic field in the pulse?Bmax=___T
1. [tex]7.95\cdot 10^6 J[/tex]
The total energy given to the cells during one pulse is given by:
[tex]E=Pt[/tex]
where
P is the average power of the pulse
t is the duration of the pulse
In this problem,
[tex]P=1.59\cdot 10^{12}W[/tex]
[tex]t=5.0 ns = 5.0\cdot 10^{-9} s[/tex]
Substituting,
[tex]E=(1.59\cdot 10^{12}W)(5.0 \cdot 10^{-6}s)=7.95\cdot 10^6 J[/tex]
2. [tex]1.26\cdot 10^{21}W/m^2[/tex]
The energy found at point (1) is the energy delivered to 100 cells. The radius of each cell is
[tex]r=\frac{4.0\mu m}{2}=2.0 \mu m = 2.0\cdot 10^{-6}m[/tex]
So the area of each cell is
[tex]A=\pi r^2 = \pi (2.0 \cdot 10^{-6}m)^2=1.26\cdot 10^{-11} m^2[/tex]
The energy is spread over 100 cells, so the total area of the cells is
[tex]A=100 (1.26\cdot 10^{-11} m^2)=1.26\cdot 10^{-9} m^2[/tex]
And so the intensity delivered is
[tex]I=\frac{P}{A}=\frac{1.59\cdot 10^{12}W}{1.26\cdot 10^{-9} m^2}=1.26\cdot 10^{21}W/m^2[/tex]
3. [tex]9.74\cdot 10^{11} V/m[/tex]
The average intensity of an electromagnetic wave is related to the maximum value of the electric field by
[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
E is the amplitude of the electric field
Solving the formula for E, we find:
[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(1.26\cdot 10^{21} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12}F/m)}}=9.74\cdot 10^{11} V/m[/tex]
4. 3247 T
The magnetic field amplitude is related to the electric field amplitude by
[tex]E=cB[/tex]
where
E is the electric field amplitude
c is the speed of light
B is the magnetic field
Solving the equation for B and substituting the value of E that we found at point 3, we find
[tex]B=\frac{E}{c}=\frac{9.74\cdot 10^{11} V/m}{3\cdot 10^8 m/s}=3247 T[/tex]
One model for a certain planet has a core of radius R and mass M surrounded by an outer shell of inner radius R, outer radius 2R, and mass 4M. If M = 2.48 × 1024 kg and R = 1.17 × 106 m, what is the gravitational acceleration of a particle at points (a) R and (b) 3R from the center of the planet?
(a) 120.8 m/s^2
The gravitational acceleration at a generic distance r from the centre of the planet is
[tex]g=\frac{GM'}{r^2}[/tex]
where
G is the gravitational constant
M' is the mass enclosed by the spherical surface of radius r
r is the distance from the centre
For this part of the problem,
[tex]r=R=1.17\cdot 10^6 m[/tex]
so the mass enclosed is just the mass of the core:
[tex]M'=M=2.48\cdot 10^{24}kg[/tex]
So the gravitational acceleration is
[tex]g=\frac{(6.67\cdot 10^{-11})(2.48\cdot 10^{24}kg)}{(1.17\cdot 10^6 m)^2}=120.8 m/s^2[/tex]
(b) 67.1 m/s^2
In this part of the problem,
[tex]r=3R=3(1.17\cdot 10^6 m)=3.51\cdot 10^6 m[/tex]
and the mass enclosed here is the sum of the mass of the core and the mass of the shell, so
[tex]M'=M+4M=5M=5(2.48\cdot 10^{24}kg)=1.24\cdot 10^{25}kg[/tex]
so the gravitational acceleration is
[tex]g=\frac{(6.67\cdot 10^{-11})(1.24\cdot 10^{25}kg)}{(3.51\cdot 10^6 m)^2}=67.1 m/s^2[/tex]
Why is the wavelike nature of a moving baseball typically NOT observed? a. The baseball's energy is too small. b. The baseball's velocity is too small. c. The baseball's wavelength is too small. d. The baseball does not have a wavelike nature. e. The baseball's frequency is too small.
(c) The baseball's wavelength is too small. Option C is correct.
What is wavelength?Wavelength is a fundamental property of waves, including electromagnetic waves like light and radio waves, as well as other types of waves, such as sound waves or water waves.
The wavelike nature of a moving baseball is typically not observed because the baseball's wavelength is too small to be measurable on a macroscopic scale.
According to the de Broglie wavelength equation, the wavelength of a moving object is inversely proportional to its momentum, which is the product of its mass and velocity. Because a baseball has a relatively large mass and a relatively low velocity compared to microscopic particles like electrons, its momentum is relatively large, which means its wavelength is extremely small, on the order of 10⁻³⁴ meters.
Thus, this is much smaller than the size of a typical baseball, so the wavelike nature of baseball is not observable in everyday situations.
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A disk with mass m = 9.5 kg and radius R = 0.3 m begins at rest and accelerates uniformly for t = 18.1 s, to a final angular speed of ω = 28 rad/s. 9) What is the angular acceleration of the disk? rad/s2 10) What is the angular displacement over the 18.1 s? rad 11) What is the moment of inertia of the disk? kg-m2 12) What is the change in rotational energy of the disk? J 13) What is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s2 14) What is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s2 15) What is the final speed of a point on the disk half-way between the center of the disk and the rim? m/s 16) What is the total distance a point on the rim of the disk travels during the 18.1 seconds? m
9) 1.55 rad/s^2
The angular acceleration of the disk is given by
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 28 rad/s[/tex] is the final angular speed
[tex]\omega_i=0[/tex] is the initial angular speed (the disk starts from rest)
t = 18.1 s is the time interval
Substituting into the equation, we find:
[tex]\alpha = \frac{28.1 rad/s - 0}{18.1 s}=1.55 rad/s^2[/tex]
10) 253.9 rad
The angular displacement of the disk during this time interval is given by the equation:
[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]
where
[tex]\omega_i=0[/tex] is the initial angular speed (the disk starts from rest)
t = 18.1 s is the time interval
[tex]\alpha=1.55 rad/s^2[/tex] is the angular acceleration
Substituting into the equation, we find:
[tex]\theta = 0 + \frac{1}{2}(1.55 rad/s^2)(18.1 s)^2=253.9 rad[/tex]
11) [tex]0.428 kg m^2[/tex]
The moment of inertia of a disk rotating about its axis is given by
[tex]I=\frac{1}{2}mR^2[/tex]
where in this case we have
m = 9.5 kg is the mass of the disk
R = 0.3 m is the radius of the disk
Substituting numbers into the equation, we find
[tex]I=\frac{1}{2}(9.5 kg)(0.3 m)^2=0.428 kg m^2[/tex]
12) 167.8 J
The rotational energy of the disk is given by
[tex]E_R = \frac{1}{2}I\omega^2[/tex]
where
[tex]I=0.428 kg m^2[/tex] is the moment of inertia
[tex]\omega[/tex] is the angular speed
At the beginning, [tex]\omega_i = 0[/tex], so the rotational energy is
[tex]E_i = \frac{1}{2}(0.428 kg m^2)(0)^2 = 0[/tex]
While at the end, the angular speed is [tex]\omega=28 rad/s[/tex], so the rotational energy is
[tex]E_f = \frac{1}{2}(0.428 kg m^2)(28 rad/s)^2=167.8 J[/tex]
So, the change in rotational energy of the disk is
[tex]\Delta E= E_f - E_i = 167.8 J - 0 = 167.8 J[/tex]
13) [tex]0.47 m/s^2[/tex]
The tangential acceleration can be found by using
[tex]a_t = \alpha r[/tex]
where
[tex]\alpha = 1.55 rad/s^2[/tex] is the angular acceleration
r is the distance of the point from the centre of the disk; since the point is on the rim,
r = R = 0.3 m
So the tangential acceleration is
[tex]a_t = (1.55 rad/s^2)(0.3 m)=0.47 m/s^2[/tex]
14) [tex]58.8 m/s^2[/tex]
The radial (centripetal acceleration) is given by
[tex]a_r = \omega^2 r[/tex]
where
[tex]\omega[/tex] is the angular speed, which is half of its final value, so
[tex]\omega=\frac{28 rad/s}{2}=14 rad/s[/tex]
r is the distance of the point from the centre (as before, r = R = 0.3 m)
Substituting numbers into the equation,
[tex]a_r = (14 rad/s)^2 (0.3 m)=58.8 m/s^2[/tex]
15) 4.2 m/s
The tangential speed is given by:
[tex]v=\omega r[/tex]
where
[tex]\omega = 28 rad/s[/tex] is the angular speed
r is the distance of the point from the centre of the disk, so since the point is half-way between the centre of the disk and the rim,
[tex]r=\frac{R}{2}=\frac{0.3 m}{2}=0.15 m[/tex]
So the tangential speed is
[tex]v=(28 rad/s)(0.15 m)=4.2 m/s[/tex]
16) 77.0 m
The total distance travelled by a point on the rim of the disk is
[tex]d=ut + \frac{1}{2}a_t t^2[/tex]
where
u = 0 is the initial tangential speed
t = 18.1 s is the time
[tex]a_t = 0.47 m/s^2[/tex] is the tangential acceleration
Substituting into the equation, we find
[tex]d=0+\frac{1}{2}(0.47 m/s^2)(18.1 s)^2=77.0 m[/tex]
A mass of 2000 kg. is raised 5.0 m in 10 seconds. What is the potential energy of the mass at this height?
98,000 J
9800 J
0
Answer:
98,000J
Explanation:
Given parameters :
Mass = 2000kg
Height = 5m
Time = 10s
P. E = mgh
P. E = 2000x 5x 9.8
P.E = 98,000J
The potential energy of a 2000 kg mass raised to a height of 5.0 m is calculated using the formula PE = mgh, resulting in a potential energy of 98000 Joules.
To determine the potential energy of a mass at a certain height, we can use the formula for gravitational potential energy, which is PE = mgh, where:
m is the mass of the objectg is the acceleration due to gravity (9.8 m/s2 on Earth)h is the height above the reference pointIn this case, the mass m is 2000 kg, g is 9.8 m/s2, and the height h is 5.0 m. So the potential energy can be calculated as follows:
PE = (2000 kg)
(9.8 m/s2)
(5.0 m) = 98000 kg
m2 s−2 = 98000 Joules
Therefore, the potential energy of the mass at 5.0 m height is 98000 J.
If potential energy at point A is 12 joules, what is the potential energy say C?
P.E. = 4
6
Or 4.8
Answer:
P.E. = 6J
Explanation:
Hope this helps. The answer above is incorrect. It's 6.
To determine the potential energy at point C, we need to know the height difference between points A and C. The potential energy at point C can be calculated using the formula P.E. = mgh. Depending on the height difference given, the potential energy at point C would be either 16 joules or 18 joules.
Explanation:To determine the potential energy at point C, we need to know the height difference between points A and C. The formula for potential energy is P.E. = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. If we assume the height difference is the same as in the given options, 4 or 6, we can calculate the potential energy at point C.
If the height difference is 4, then the potential energy at C would be 12 joules + 4 joules = 16 joules. If the height difference is 6, then the potential energy at C would be 12 joules + 6 joules = 18 joules.
Therefore, the potential energy at point C is either 16 joules or 18 joules, depending on the height difference between points A and C.
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Two vectors are illustrated in the coordinate plane. What are the components of the vector in quadrant I?
Answer:
option D
(2.6 , 3.1 )
Explanation:
First quadrant is where x and y values are positive.
Given in the question,
magnitude of vector = 4
angle at which vector is incline from x-axis = 50°
Vertical component (y component) sinΔ = opp / hyposin(50) = vertical / 4
vertical = sin(50)(4)
= 3.06
≈ 3.1
Horizontal component (x component)cosΔ = adj / hypo
cos(50) = horizontal / 4
horizontal = cos(50)(4)
= 2.57
≈ 2.6
The Global Positioning System (GPS) network consists of 24 satellites, each of which makes two orbits around the Earth per day. Each satellite transmits a 50.0-W sinusoidal electromagnetic signal at two frequencies, one of which is 1575.42 MHz. Assume that a satellite transmits half of its power at each frequency, and that the waves travel uniformly in a downward hemisphere. a. What average intensity does a GPS receiver on the ground, directly below the satellite, receive? (Hint: for satellites in Earth orbit, the gravitational force supplies the centripetal force for the satellite to remain in orbit). b. What are the amplitudes of the electric and magnetic fields at the GPS receiver in part (a), and how long does it take the signal to reach the receiver? c. If the receiver is a square panel 1.50 cm on a side that absorbs all of the beam, what average pressure does the signal exert on it? d. What wavelength must the receiver be tuned to?
The average intensity, electric and magnetic field amplitudes at the GPS receiver are determined based on the allocated power and the area over which it is distributed. The signal pressure on the receiver and the time it takes for the signal to reach the receiver are also calculated. The wavelength for receiver tuning is derived from the transmitted frequency.
When analyzing the Global Positioning System (GPS), the average intensity received by a GPS receiver on the ground directly below the satellite can be calculated by considering the satellite's power output and the area over which the power is distributed. Since each satellite transmits at two frequencies with a combined power of 50.0 W and the power is transmitted uniformly in a downward hemisphere, the intensity can be calculated by dividing the power allocated to one frequency (25 W) by the surface area of the hemisphere over which it spreads. The area of a hemisphere is given by 2πr², where r is the distance from the satellite to the receiver on Earth's surface.
The electric and magnetic field amplitudes can be found using the intensity and the relationships inherent in electromagnetic waves. To calculate the pressure the signal exerts on a square panel GPS receiver, we can use the relationship between intensity, pressure, and the speed of light. For finding the wavelength that the receiver must be tuned to, we use the formula λ = c / f, where c is the speed of light and f is the frequency of the transmitted signal.
The answer to part (b) also includes the time it takes for the signal to reach the receiver. This can be calculated by dividing the distance of the satellite from the Earth's surface by the speed of light, considering that electromagnetic signals travel at the speed of light in a vacuum.
Finally, for part (d), the given frequency of 1575.42 MHz is used to determine the wavelength of the signal. We can ignore any potential distortions or delays that would alter the frequency and its corresponding wavelength at the receiver level under normal GPS operational conditions.
(a) The average intensity is [tex]I=9.66\times 10^{-15} W/m^2[/tex]. (b) The amplitude of the electric and magnetic fields is [tex]E=2.70\times 10^{-6} V/m[/tex] and [tex]B=9.0\times 10^{-15}T[/tex], and the signal takes the time is [tex]t=0.068 s[/tex]. (c) The average pressure is [tex]p=3.22\times 10^{-23}Pa[/tex]. (d) The wavelength of the signal is [tex]\lambda=0.190m[/tex].
(a)
The gravitational attraction between the Earth and the satellite is equal to the centripetal force that keeps the satellite in circular motion, So
[tex]\frac{GmM}{r^2} = m\omega^2 r[/tex]
Or [tex]r=\sqrt[3]{\frac{GM}{\omega^2} }[/tex] ...(1)
where
[tex]G[/tex] is the gravitational constant
[tex]m[/tex] is the satellite's mass
[tex]M[/tex] is the earth's mass
[tex]r[/tex] is the distance of the satellite from the Earth's center
[tex]\omega[/tex] is the angular frequency of the satellite
The satellite here makes two orbits around the Earth per day, So its frequency is
[tex]\omega = \frac {2 (rev)/(day)}{24 (h)/(day) \times 60 (min)/(h)} \times \frac {(2\pi rad/rev)}{(s/min)}=1.45\times 10^{-4} rad/s[/tex]
By solving equation (1),
[tex]r=\sqrt[3]{\frac{GM}{\omega^2} }=\sqrt[3]{(\frac{6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)}{(1.45\cdot 10^{-4} rad/s)^2}}=2.67\times 10^7 m[/tex]
The radius of the Earth is
[tex]R=6.37\times 10^6 m[/tex]
So, the altitude of the satellite is
[tex]h=r-R=2.67\times 10^7 m-6.37\times 10^6m=2.03\times 10^7 m[/tex]
The average intensity received by a GPS receiver on the Earth will be given by
[tex]I=\frac{P}{A}[/tex]
where
[tex]P[/tex] is the power given as [tex]P=50 W[/tex]
[tex]A[/tex] is the area of a hemisphere given as [tex]A=4\pi h^2 = 4 \pi (2.03\times 10^7 m)^2=5.18\times 10^{15} m^2[/tex]
Now the intensity is given by
[tex]I=\frac{50.0 W}{5.18\times 10^{15}m^2}=9.66\times 10^{-15} W/m^2[/tex]
(b)
The relationship between average intensity of an electromagnetic wave and amplitude of the electric field is
[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]
where
[tex]c[/tex] is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
[tex]E[/tex] is the amplitude of the electric field
The amplitude of the electric field is given by
[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(9.66\times 10^{-15} W/m^2)}{(3\times 10^8 m/s)(8.85\times 10^{-12}F/m))}}=2.70\times 10^{-6} V/m[/tex]
The amplitude of the magnetic field is given by
[tex]B=\frac{E}{c}=\frac{2.70\times 10^{-6} V/m}{3\times 10^8 m/s}=9.0\times 10^{-15}T[/tex]
The signal travels at the speed of light, so the time it takes to reach the Earth is the distance covered divided by the speed of light:
[tex]t=\frac{h}{c}=\frac{2.03\times 10^7 m}{3\times 10^8 m/s}=0.068 s[/tex]
(c)
In the case of a perfect absorber, the radiation pressure exerted by an electromagnetic wave on a surface is given by
[tex]p=\frac{I}{c}[/tex]
where
[tex]I[/tex] is the average intensity
[tex]c[/tex] is the speed of light
Plugging the values
[tex]I=9.66\times 10^{-15} W/m^2[/tex]
So the average pressure is
[tex]p=\frac{9.66\times 10^{-15} W/m^2}{3\times 10^8 m/s}=3.22\times 10^{-23}Pa[/tex]
(d)
The wavelength of the receiver must be tuned to the same wavelength as the transmitter (the satellite), which is given by
[tex]\lambda=\frac{c}{f}[/tex]
where
[tex]c[/tex] is the speed of light
[tex]f[/tex] is the frequency of the signal
For the satellite in the problem, the frequency is
[tex]f=1575.42 MHz=1575.42\times 10^6 Hz[/tex]
So the wavelength of the signal is
[tex]\lambda=\frac{3.0\times 10^8 m/s}{1575.42 \times 10^6 Hz}=0.190 m[/tex]
A metal detector uses a changing magnetic field to detect metallic objects. Suppose a metal detector that generates a uniform magnetic field perpendicular to its surface is held stationary at an angle of 15.0∘ to the ground, while just below the surface there lies a silver bracelet consisting of 6 circular loops of radius 5.00 cm with the plane of the loops parallel to the ground. If the magnetic field increases at a constant rate of 0.0250 T/s, what is the induced emf E? Take the magnetic flux through an area to be positive when B⃗ crosses the area from top to bottom.
Answer:
[tex]-1.14 \cdot 10^{-3} V[/tex]
Explanation:
The induced emf in the loop is given by Faraday's Newmann Lenz law:
[tex]\epsilon = - \frac{d \Phi}{dt}[/tex] (1)
where
[tex]d\Phi[/tex] is the variation of magnetic flux
[tex]dt[/tex] is the variation of time
The magnetic flux through the coil is given by
[tex]\Phi = NBA cos \theta[/tex] (2)
where
N = 6 is the number of loops
A is the area of each loop
B is the magnetic field strength
[tex]\theta =15^{\circ}[/tex] is the angle between the direction of the magnetic field and the normal to the area of the coil
Since the radius of each loop is r = 5.00 cm = 0.05 m, the area is
[tex]A=\pi r^2 = \pi (0.05 m)^2=0.0079 m^2[/tex]
Substituting (2) into (1), we find
[tex]\epsilon = - \frac{d (NBA cos \theta)}{dt}= -(NAcos \theta) \frac{dB}{dt}[/tex]
where
[tex]\frac{dB}{dt}=0.0250 T/s[/tex] is the rate of variation of the magnetic field
Substituting numbers into the last formula, we find
[tex]\epsilon = -(6)(0.0079 m^2)(cos 15^{\circ})(0.0250 T/s)=-1.14 \cdot 10^{-3} V[/tex]
Answer:
Induced emf, [tex]\epsilon=-1.13\times 10^{-3}\ volts[/tex]
Explanation:
It is given that,
Number of circular loop, N = 6
A uniform magnetic field perpendicular to its surface is held stationary at an angle of 15 degrees to the ground.
Radius of the loop, r = 5 cm = 0.05 m
Change in magnetic field, [tex]\dfrac{dB}{dt}=0.025\ T/s[/tex]
Due to the change in magnetic field, an emf will be induced. Let E is the induced emf in the coil. it is given by :
[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]
[tex]\phi[/tex] = magnetic flux
[tex]\epsilon=\dfrac{d(NBA\ cos\theta)}{dt}[/tex]
[tex]\epsilon=-NA\dfrac{d(B)}{dt}[/tex]
[tex]\epsilon=6\times \pi (0.05)^2\times 0.025\times cos(15)[/tex]
[tex]\epsilon=-1.13\times 10^{-3}\ volts[/tex]
So, the induced emf in the loop is [tex]1.13\times 10^{-3}\ volts[/tex] . Hence, this is the required solution.
Nick’s swimming goggles are made of blue plastic. When he straps them on, he notices that the goggles change the color of the objects around him. A white ring buoy appears because the blue plastic all colors of light except blue. Only the blue light the ring buoy passes through the blue plastic.
A white ring buoy appears blue because the blue plastic absorbs all colors of light except blue. Only the blue light reflected from the ring buoy passes through the blue plastic.
Answer:
blue
absorbs
reflected from
Explanation:
I looked it up and everybody said this :)
At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does the potential difference across the resistor compare to the emf across the coil? (Enter your answers in V.) resistor V coil V (b) Answer the same question about the circuit several seconds later. (Enter your answers in V.) resistor V coil V (c) Is there an instant at which these two voltages are equal in magnitude? Yes No (d) If so, when? Is there more than one such instant? (Enter all possible times in ms as a comma-separated list. If there are no such instants, enter NONE.) ms (e) After a 3.20 A current is established in the resistor and coil, the battery is suddenly replaced by a short circuit. Answer questions (a) and (b) again with reference to this new circuit. (Enter your answers in V.) immediately thereafter several seconds later resistor V V Need Help?
(a) On the coil: 20 V, on the resistor: 0 V
The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:
[tex]V = V_R + V_L[/tex]
The potential difference across the inductance is given by
[tex]V_L(t) = V e^{-\frac{t}{\tau}}[/tex] (1)
where
[tex]\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s[/tex] is the time constant of the circuit
At time t=0,
[tex]V_L(0) = V e^0 = V = 20 V[/tex]
So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:
[tex]V_R = V-V_L = 20 V-20 V=0[/tex]
(b) On the coil: 0 V, on the resistor: 20 V
Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for
[tex]t >> \tau[/tex]
Since [tex]\tau[/tex] is at the order of less than milliseconds.
Using eq.(1), we see that for [tex]t >> \tau[/tex], the exponential becomes zero, and therefore the potential difference across the coil is zero:
[tex]V_L = 0[/tex]
Therefore, the potential difference across the resistor will be
[tex]V_R = V-V_L = 20 V- 0 = 20 V[/tex]
(c) Yes
The two voltages will be equal when:
[tex]V_L = V_R [/tex] (2)
Reminding also that the sum of the two voltages must be equal to the voltage of the battery:
[tex]V=V_L +V_R[/tex]
And rewriting this equation,
[tex]V_R = V-V_L[/tex]
Substituting into (2) we find
[tex]V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V[/tex]
So, the two voltages will be equal when they are both equal to 10 V.
(d) at [tex]t=5.77\cdot 10^{-4}s[/tex]
We said that the two voltages will be equal when
[tex]V_L=\frac{V}{2}[/tex]
Using eq.(1), and this last equation, this means
[tex]V e^{-\frac{t}{\tau}} = \frac{V}{2}[/tex]
And solving the equation for t, we find the time t at which the two voltages are equal:
[tex]e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s[/tex]
(e-a) -19.2 V on the coil, 19.2 V on the resistor
Here we have that the current in the circuit is
[tex]I_0 = 3.20 A[/tex]
The problem says this current is stable: this means that we are in a situation in which [tex]t>>\tau[/tex], so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law
[tex]V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V[/tex]
Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.
The voltage on the coil is given by
[tex]V_L(t) = -V e^{-\frac{t}{\tau}}[/tex] (1)
which means that it is maximum at the moment when the battery is disconnected, when t=0:
[tex]V_L(0)=.V[/tex]
And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.
(e-b) 0 V on both
After several seconds,
[tex]t>>\tau[/tex]
If we use this approximation into the formula
[tex]V_L(t) = -V e^{-\frac{t}{\tau}}[/tex] (1)
We find that
[tex]V_L = 0[/tex]
And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.
When a battery is connected to a coil and a resistor, initially all the battery voltage is across the coil. After some time, the voltage across the coil reduces to zero and that across the resistor becomes equal to the battery voltage. The voltages are equal during the magnetization process of the coil. When the battery is replaced with a short circuit, the entire voltage falls across the coil initially and then reduces to zero.
Explanation:At the instant the battery is connected to the 5.00mH coil and 6.00Ω resistor (t=0), the potential difference across the resistor will be zero and the entire battery voltage will be across the coil as it opposes the sudden change in current. Therefore, at t=0, the voltage across the resistor is 0V and the voltage across the coil is 20.0V.
After several seconds, the current in the circuit will have achieved a steady state as the coil has become fully magnetized and now behaves as a wire. Therefore, the voltage drop across the resistor will become 20.0V (due to Ohm's law I=V/R, V=IR), and that across the coil will be 0V.
The two voltages are equal when the coil is half-way through its magnetization process. This is when the current in the circuit is building up.
When the circuit is modified with a short circuit after a current of 3.20A has been established, the entire potential difference falls across the coil immediately afterwards. Several seconds later, the coil loses its magnetic field since no more current flows through the circuit, thus no voltage exists either across the coil or the resistor.
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A constant force is applied to an object, causing the object to accelerate at 8 m/s2. What would the acceleration be if each of the following things happened (from the initial state)? (a) the force is doubled (b) the object's mass is doubled (c) the force and the object's mass are both doubled (d) the force is doubled and the object's mass is halved
Final answer:
Acceleration calculations for different scenarios with varying forces and masses applied to an object initially accelerating at 8 m/s².
Explanation:
Acceleration calculation:
Initial scenario: Force = 6 N, mass = 24 kg, resulting in acceleration = 6 N / 24 kg = 0.25 m/s².(a) Force doubled: New force = 12 N, with the same mass, acceleration = 12 N / 24 kg = 0.5 m/s².(b) Mass halved: New mass = 12 kg, with the force remaining the same, acceleration = 6 N / 12 kg = 0.5 m/s².(c) Force and mass doubled: Force = 12 N, mass = 48 kg, so acceleration = 12 N / 48 kg = 0.25 m/s².(d) Force doubled and mass halved: Force = 12 N, mass = 12 kg, acceleration = 12 N/12 kg = 1 m/s². What is the best description of the destructive interference of light?
A. A mechanical wave meets an electromagnetic wave
B. Two waves have displaced in opposite directions
C.two waves have the same direction of displacement
D. A longitudinal wave meets a transverse light wave
(It is B)
Answer:
B. Two waves have displaced in opposite directions
Explanation:
Interference occurs when two waves meet at a point in space. When this occurs, two extreme conditions can occur:
- if the two waves are in phase (=displacement in the same direction), the amplitude of the resultant wave is equal to the sum of the amplitudes of the two waves:
A = A1 + A2
and this condition is called constructive interference
- if the two waves are in anti-phase (=displacement in opposite directions), the amplitude of the resultant wave is equal to the difference of the amplitudes of the two waves:
A = |A1 - A2|
and this condition is called destructive interference. Note that if A1=A2, the amplitude of the resultant wave is zero.
A fixed system of charges exerts a force of magnitude 22 N on a 4.0 C charge. The 4.0 C charge is replaced with a 8.0 C charge. What is the exact magnitude of the force (in N) exerted by the system of charges on the 8.0 C charge? Do not include units with your answer.
Answer:
44 N
Explanation:
The electrostatic forces between two charges is given by:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is their separation
We notice that the force is directly proportional to the charges.
In this problem, initially we have a force of
F = 22 N
on a q2 = 4.0 C, exerted by a charge q1.
If the charge is doubled,
q2 = 8.0 C
This means that the force will also double, so it will be
[tex]F=22 N \cdot 2 = 44 N[/tex]
Final answer:
The force exerted by the fixed system of charges on the 8.0 C charge will be double the force exerted on the 4.0 C charge, resulting in a force of 44 N.
Explanation:
The force exerted on a charge by a system of charges is directly proportional to the charge itself. Hence, when we replace the 4.0 C charge in the initial system with an 8.0 C charge, we can determine the new force by recognizing that the charged system has simply doubled. If the 4.0 C charge experiences 22 N of force, we can calculate the force on the 8.0 C charge by multiplying the force by two (since 8.0 C is twice 4.0 C).
F_{new} = 2 \times F_{old}
F_{new} = 2 \times 22 \ N
F_{new} = 44 \ N
The force on the 8.0 C charge would therefore be 44 N.
The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail. When small prey, or even potential mates, swim within a few feet of the fish, they perturb the electric field. The change in the field is picked up by electric sensor cells in the skin of the elephant nose. These remarkable fish can detect changes in the electric field as small as 3.00 μN/C. How much charge, modeled as a point charge, in the fish would be needed to produce such a change in the electric field at a distance of 63.5 cm ?
Answer:
[tex]1.34\cdot 10^{-16} C[/tex]
Explanation:
The strength of the electric field produced by a charge Q is given by
[tex]E=k\frac{Q}{r^2}[/tex]
where
Q is the charge
r is the distance from the charge
k is the Coulomb's constant
In this problem, the electric field that can be detected by the fish is
[tex]E=3.00 \mu N/C = 3.00\cdot 10^{-6}N/C[/tex]
and the fish can detect the electric field at a distance of
[tex]r=63.5 cm = 0.635 m[/tex]
Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:
[tex]Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^{-6} N/C)(0.635 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=1.34\cdot 10^{-16} C[/tex]
Final answer:
A point charge of approximately 1.35 × 10-13 coulombs would be required to produce an electric field change of 3.00 µN/C at a distance of 63.5 cm from the elephant nose fish.
Explanation:
To calculate the amount of charge needed to produce a change in the electric field of 3.00 µN/C at a distance of 63.5 cm, we can use Coulomb's law, which describes the electric field E created by a point charge Q. Coulomb's law is given by the equation E = kQ/r2, where k is Coulomb's constant (8.987 × 109 N m2/C2), Q is the charge in coulombs, and r is the distance from the charge in meters. To find Q, we rearrange the equation to Q = Er2/k.
By substituting the values, E = 3.00 µN/C (or 3.00 × 10-6 N/C), and the distance r = 63.5 cm (or 0.635 m), we can calculate the charge Q necessary to produce the observed electric field at the specified distance.
Calculation:
Q = (3.00 × 10-6 N/C) × (0.635 m)2 / (8.987 × 109 N m2/C2)
= 1.35 × 10-13 C
So, a point charge of approximately 1.35 × 10-13 coulombs would be needed to produce an electric field change of 3.00 µN/C at a distance of 63.5 cm from the elephant nose fish.
Nitrogen is used in many plant fertilizers. When it rains, the excess fertilizers run off into streams, rivers, and eventually are carried to the ocean. Algae in the water feed off of the nitrogen in the fertilizers. The excess nutrients can cause a huge mat of algae to grow, called an algal bloom.
When the nitrogen is all used up, the algae die and decomposing bacteria begin to consume them. The bacteria multiply and consume most of the oxygen in the water, causing many of the fish and other larger organisms in the water to suffocate and die.
How could the problem of algal blooms be solved?
A.
The government should start a program in which they clean out all the algae from the lakes and streams before they become a problem.
B.
Antibiotics should be added to the oceans to kill all of the bacteria before they can consume the dead algae.
C.
Farmers and gardeners should use other methods to provide crops with the nitrogen they need.
D.
Fish should be trained to swim to other parts of a body of water when the oxygen levels in one area become too low.
Answer:
C. Farmers and gardeners should use other methods to provide crops with the nitrogen they need.
Explanation:
That was the right answer on study island :/
Coherent red light of wavelength λ = 700 nm is incident on two very narrow slits. The light has the same phase at both slits. (a) What is the angular separation in radians between the central maximum in intensity and an adjacent maximum if the slits are 0.025 mm apart? (b) What would be the separation between maxima in intensity on a screen located 1 m from the slits? (c) What would the angular separation be if the slits were 2.5 mm apart? (d) What would the separation between maxima be on a screen 25 mm from the slits? The answer is relevant to the maximum resolution along your retina that would be useful given the size of your pupil. The spacing between cones on the retina is about 10 µm. (e) How would the location of the maxima change if one slit was covered by a thin film with higher index of refraction that shifted the phase of light leaving the slit by π? Would the separation change?
(a) 0.028 rad
The angular separation of the nth-maximum from the central maximum in a diffraction from two slits is given by
[tex]d sin \theta = n \lambda[/tex]
where
d is the distance between the two slits
[tex]\theta[/tex] is the angular separation
n is the order of the maximum
[tex]\lambda[/tex] is the wavelength
In this problem,
[tex]\lambda=700 nm=7\cdot 10^{-7} m[/tex]
[tex]d=0.025 mm=2.5\cdot 10^{-5} m[/tex]
The maximum adjacent to the central maximum is the one with n=1, so substituting into the formula we find
[tex]sin \theta = \frac{n \lambda}{d}=\frac{(1)(7\cdot 10^{-7} m)}{2.5\cdot 10^{-5} m}=0.028[/tex]
So the angular separation in radians is
[tex]\theta= sin^{-1} (0.028) = 0.028 rad[/tex]
(b) 0.028 m
The screen is located 1 m from the slits:
D = 1 m
The distance of the screen from the slits, D, and the separation between the two adjacent maxima on the screen (let's call it y) form a right triangle, so we can write the following relationship:
[tex]\frac{y}{D}=tan \theta[/tex]
And so we can find y:
[tex]y=D tan \theta = (1 m) tan (0.028 rad)=0.028 m[/tex]
(c) [tex]2.8\cdot 10^{-4} rad[/tex]
In this case, we can apply again the formula used in part a), but this time the separation between the slits is
[tex]d=2.5 mm = 0.0025 m[/tex]
so we find
[tex]sin \theta = \frac{n \lambda}{d}=\frac{(1)(7\cdot 10^{-7} m)}{0.0025 m}=2.8\cdot 10^{-4}[/tex]
And so we find
[tex]\theta= sin^{-1} (2.8\cdot 10^{-4}) = 2.8\cdot 10^{-4} rad[/tex]
(d) [tex]7.0\cdot 10^{-6} m = 7.0 \mu m[/tex]
This part can be solved exactly as part b), but this time the distance of the screen from the slits is
[tex]D=25 mm=0.025 m[/tex]
So we find
[tex]y=D tan \theta = (0.025 m) tan (2.8\cdot 10^{-4} rad)=7.0\cdot 10^{-6} m = 7.0 \mu m[/tex]
(e) The maxima will be shifted, but the separation would remain the same
In this situation, the waves emitted by one of the slits are shifted by [tex]\pi[/tex] (which corresponds to half a cycle, so half wavelength) with respect to the waves emitted by the other slit.
This means that the points where previously there was constructive interference (the maxima on the screen) will now be points of destructive interference (dark fringes); on the contrary, the points where there was destructive interference before (dark fringes) will now be points of maxima (bright fringes). Therefore, all the maxima will be shifted.
However, the separation between two adjacent maxima will not change. In fact, tall the maxima will change location exactly by the same amount; therefore, their relative distance will remain the same.
Which process helps regulate Earth's climate by transporting warm seawater to colder regions of seawater?
A. Thermohaline circulation
B. Surface tension
C. Centrifugal force
D. The tides
Answer: A. Thermohaline circulation
Explanation:
The thermohaline circulation is a phenomena which involves the movement of the oceanic currents because of the differences that occur in the temperature and salinity of oceanic water. These two factors changes the salinity and density of the water and fluctuating the climatic conditions. Cold water is particularly denser than the warm water. The water with more salinity is also denser over water with less salinity. The deep oceanic water currents brings changes in the density as well as the salinity of water hence, the circulation of water is called as thermohaline circulation. Some part of warm water will evaporate from the surface layer of the ocean this will lead to the flow of warm wind currents and water to the regions of cold regions of the seawater.
On the basis of the above description, A. Thermohaline circulation is the correct option.
The correct answer is A. Thermohaline circulation
Explanation:
Thermohaline circulation is a complex natural phenomenon that involves the flow of water with different temperatures, density and even salinity in oceans all around the world. This circulation of water plays an important role in the climate as warm seawater flows to regions with colder seawater. For example, warm seawater from tropical regions flows to polar regions, and this warms polar regions and it is responsible for regulating ice formation. According to this, it is Thermohaline circulation the process that helps regulate Earth's climate by transporting warm seawater to colder regions of seawater.
The nucleus of an atom consists of protons and neutrons (no electrons). A nucleus of a carbon‑12 isotope contains six protons and six neutrons, while a nitrogen‑14 nucleus comprises seven protons and seven neutrons. A graduate student performs a nuclear physics experiment in which she bombards nitrogen‑14 nuclei with very high speed carbon‑12 nuclei emerging from a particle accelerator. As a result of each such collision, the two nuclei disintegrate completely, and a mix of different particles are emitted, including electrons, protons, antiprotons (with electric charge −???? each), positrons (with charge +???? each), and various neutral particles (including neutrons and neutrinos). For a particular collision, she detects the emitted products and find 17 protons, 4 antiprotons, 7 positrons, and 25 neutral particles. How many electrons are also emitted?
Answer:
7 electrons
Explanation:
We can solve the problem by using the law of conservation of electric charge: in fact, the total electric charge before and after the collision must be conserved.
Before the collision, we have:
- A nucleus of carbon-12, consisting of 6 protons (charge +1 each) + 6 neutrons (charge 0 each), so total charge of +6
- A nucleus of nitrogen-14, consisting of 7 protons (charge +1 each) + 7 neutrons (charge 0 each), so total charge of +7
So the total charge before the collision is +6+7=+13 (1)
After the collision, we have:
- 17 protons (charge +1 each): total charge of +17
- 4 antiprotons (charge -1 each): total charge of -4
- 7 positrons (charge +1 each): total charge of +7
- 25 neutral particles (charge 0 each): total charge of 0
- N electrons (charge -1 each): total charge of -N
So the total charge after the collision is +17-4+7+0-N=+20-N (2)
Since the charge must be conserved, we have (1) = (2):
+13 = +20 - N
Solving for N,
N = 20 - 13 = 7
So, there are 7 electrons.
The number of electrons emitted in the collision between carbon-12 and nitrogen-14 nuclei detected as 17 protons, 4 antiprotons, 7 positrons, and 25 neutral particles, can be calculated to be 20 to maintain charge neutrality.
Explanation:When the graduate student bombards nitrogen-14 nuclei with carbon-12 nuclei and detects the disintegration products including 17 protons, 4 antiprotons, 7 positrons, and 25 neutral particles, the amount of emitted electrons can be calculated by considering the charge balance of the particles. Since protons have a +1 charge and electrons have a -1 charge, positrons also have a +1 charge and antiprotons have a -1 charge, the net charge must remain the same as before the collision.
Originally, the nitrogen-14 nucleus (with 7 protons) and the carbon-12 nucleus (with 6 protons) totaled 13 positive charges. After the collision, the detected 17 protons and 7 positrons contribute +24 to the net charge, while 4 antiprotons contribute -4, making the total positive charge +20. To counterbalance this and maintain a neutral charge, 7 additional electrons (on top of the original 13) must have been emitted to bring the net charge back down to +13. Thus, in total, there must be 20 electrons emitted.
If the area is 2 square meters (2 m²), what's the force of gravity acting on the column of water?
What area ?
What column of water ?
The force of gravity acting on a column of water with an area of 2 square meters and height of 1 meter is 19620 Newtons. This calculation is based on the assumption of the density of water being 1000 kg/m³ and the acceleration due to gravity being 9.81 m/s².
Explanation:The force of gravity acting on a column of water with an area of 2 square meters can be calculated using the concepts of fluid pressure and weight. First, you need to understand that the force of gravity on the water can be expressed by the formula Weight = Mass x Gravity. The Mass can be derived from the Volume (Area x Height) and the density of the water (approximately 1000 kg/m³).
Thus, assuming the column of water is 1m high, the volume of water is 2 m² x 1 m = 2 m³. The mass is then Volume x Density = 2 m³ x 1000 kg/m³ = 2000 kg. Substituting these values into the weight formula gives us Weight = 2000 kg x 9.81 m/s² (acceleration due to gravity), which equals 19,620 Newtons.
Therefore, the force of gravity acting on the column of water is 19620 N.
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A mass is suspended on a vertical spring. Initially, the mass is in equilibrium. Then, it is pulled downward and released. The mass then moves up and down between the "top" and the "bottom" positions. By definition, the period of such motion is the time interval it takes the mass to move: Mark all the correct statements among those provided below. View Available Hint(s) Mark all the correct statements among those provided below. from the top position to the bottom. from the equilibrium position to the bottom. from the bottom position to the top. from the equilibrium position to the bottom and then back to the equilibrium. from the equilibrium position to the top and then back to the equilibrium. from the equilibrium position to the top. from the top position to the bottom and then back to the top. from the bottom position to the top and then back to the bottom.
Answer:
1. From top to bottom and then back to top
2. From bottom to top and then back to bottom
Explanation:
As Time Period or periodic time period is time it takes to complete one complete cycle. So only these two options are correct. Yes ! If you assume a frictionless and isolated system then these two time intervals must be equal.
The motion of the suspended mass is simple harmonic motion; from the bottom position, then to equilibrium position, and then to the top position.
Period of simple harmonic motionThe period of a particle undergoing simple harmonic motion is defined as the time taken for the particle to complete one complete oscillation.
Motion of the vibrating body
A mass suspended on a vertical spring and allowed to attain equilibrium. When, it is pulled downward and released, the mass begins to oscillate by moving up and down between the "top" and the "bottom" position.
The motion of the object is as follows:
It goes to the bottom position.then to equilibrium position,then to the top positionThe motion of the suspended mass is simple harmonic motion; from the bottom position, then to equilibrium position, and then to the top position.
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When a particle of charge q moves with a velocity v⃗ in a magnetic field B⃗ , the particle is acted upon by a force F⃗ exerted by the magnetic field. To find the direction and magnitude of this force, follow the steps in the following Tactics Box. Keep in mind that the right-hand rule for forces shown in step 2 gives the direction of the force on a positive charge. For a negative charge, the force will be in the opposite direction.If the magnetic field of the wire is 4.0×10−4 T and the electron moves at 6.0×106 m/s , what is the magnitude F of the force exerted on the electron?Express your answer in newtons to two significant figures.
Answer:
[tex]3.8\cdot 10^{-16}N[/tex]
Explanation:
For a charged particle moving perpendicularly to a magnetic field, the magnitude of the force exerted on the particle is:
[tex]F=qvB[/tex]
where
q is the magnitude of the charge of the particle
v is the velocity of the particle
B is the magnetic field strength
In this problem, we have
[tex]q=e=1.6\cdot 10^{-19}[/tex] is the charge of the electron
[tex]v=6.0\cdot 10^6 m/s[/tex] is the velocity
[tex]B=4.0\cdot 10^{-4}T[/tex] is the magnetic field
Substituting into the formula, we find the force:
[tex]F=(1.6\cdot 10^{-19} C)(6.0\cdot 10^6 m/s)(4.0\cdot 10^{-4}T)=3.8\cdot 10^{-16}N[/tex]
The force exerted on the electron by the magnetic field, calculated using F = qvB sin θ, is 3.84 x 10^-16 N.
Explanation:The magnitude of the force exerted on a charged particle moving in a magnetic field can be calculated using the formula F = qvB sin θ, where F is the force, q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and magnetic field vectors. In this case with an electron moving in a straight path within a magnetic field, the angle is 90 degrees, so sin θ equals 1. The charge of an electron (q) is 1.6 x 10-19 C (Coulombs).
Therefore, substituting these values into the equation gives: F = (1.6 x 10-19 C) * (6.0 x 106 m/s) * (4.0 x 10-4 T) * 1 = 3.84 x 10-16 N (Newtons).
This means that the magnitude of the force exerted on the electron by the magnetic field is 3.84 x 10-16 N.
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Which of the following will increase the resistance of a wire?a) Decreasing the resistivity of the material the wire is composed of will increase the resistance of the wire.b) Decreasing the cross-sectional area of the wire will increase the resistance of the wire.c) Decreasing the length of the wire will increase the resistance of the wire.d) Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.e) Increasing the cross-sectional area of the wire will increase the resistance of the wire.f) Increasing the length of the wire will increase the resistance of the wire.
Answer:
b) Decreasing the cross-sectional area of the wire will increase the resistance of the wire.
d) Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.
f) Increasing the length of the wire will increase the resistance of the wire.
A horizontal spring attached to a wall has a force constant of k = 860 N/m. A block of mass m = 1.60 kg is attached to the spring and rests on a frictionless, horizontal surface as in the figure below. The left end of a horizontal spring is attached to a vertical wall, and the right end is attached to a block of mass m. The spring has force constant k. Three positions are labeled along the spring, and the block is pulled to the rightmost position, stretching the spring. The leftmost position, the equilibrium position, is labeled x = 0. The middle position, halfway between the leftmost and rightmost positions, is labeled x = xi⁄2. The rightmost position is labeled x = xi. (a) The block is pulled to a position xi = 5.00 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 5.00 cm from equilibrium. J (b) Find the speed of the block as it passes through the equilibrium position. m/s (c) What is the speed of the block when it is at a position xi/2 = 2.50 cm? m/s
(a) 1.08 J
The elastic potential energy stored in the block at any position x is given by
[tex]U=\frac{1}{2}kx^2[/tex]
where
k is the spring constant
x is the displacement relative to the equilibrium position
Here we have
k = 860 N/m
x = 5.00 cm = 0.05 m is the position of the block
Substituting, we find
[tex]U=\frac{1}{2}(860 N/m)(0.05 m)^2=1.08 J[/tex]
(b) 1.16 m/s
The total mechanical energy of the spring-mass system is equal to the potential energy found at point (a), because there the system was at its maximum displacement, where the kinetic energy (because the speed is zero).
At the equilibrium position, the mechanical energy is sum of kinetic and potential energy
E = K + U
However, at equilibrium position x = 0, so U = 0. Therefore, the kinetic energy is equal to the total energy found at point (a)
[tex]E=K= \frac{1}{2}mv^2 = 1.08 J[/tex]
where
m = 1.60 kg is the mass of the block
v is the speed
Solving for v, we find
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.08 J)}{1.60 kg}}=1.16 m/s[/tex]
(c) 1.00 m/s
When the block is at position x = 2.50 cm, the mechanical energy is sum of kinetic and potential energy:
[tex]E=K+U=\frac{1}{2}mv^2 + \frac{1}{2}kx^2[/tex]
where
E = 1.08 J is the total mechanical energy
m = 1.60 kg is the mass
v is the speed
k = 860 N/m
x = 2.50 cm = 0.025 m is the displacement
Solving for v, we find
[tex]v = \sqrt{\frac{2E - kx^2}{m}}=\sqrt{\frac{2(1.08 J)-(860 N/m)(0.025 m)^2}{1.60 kg}}=1.00 m/s[/tex]
Answer:
b
Explanation:
edg 2021