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 compact, dense object with a mass of 2.90 kg is attached to a spring and is able to oscillate horizontally with negligible friction. The object is pulled to a distance of 0.200 m from its equilibrium position, held in place with a force of 16.0 N, and then released from rest. It then oscillates in simple harmonic motion. (The object oscillates along the x-axis, where x = 0 is the equilibrium position.) (a) What is the spring constant (in N/m)? N/m (b) What is the frequency of the oscillations (in Hz)? Hz (c) What is the maximum speed of the object (in m/s)? m/s (d) At what position(s) (in m) on the x-axis does the maximum speed occur? x = ± m (e) What is the maximum acceleration of the object? (Enter the magnitude in m/s2.) m/s2 (f) At what position(s) (in m) on the x-axis does the maximum acceleration occur? x = ± m (g) What is the total mechanical energy of the oscillating spring–object system (in J)? J (h) What is the speed of the object (in m/s) when its position is equal to one-third of the maximum displacement from equilibrium? m/s (i) What is the magnitude of the acceleration of the object (in m/s2) when its position is equal to one-third of the maximum displacement from equilibrium? m/s2
(a) 80 N/m
The spring constant can be found by using Hooke's law:
[tex]F=kx[/tex]
where
F is the force on the spring
k is the spring constant
x is the displacement of the spring relative to the equilibrium position
At the beginning, we have
F = 16.0 N is the force applied
x = 0.200 m is the displacement from the equilibrium position
Solving the formula for k, we find
[tex]k=\frac{F}{m}=\frac{16.0 N}{0.200 m}=80 N/m[/tex]
(b) 0.84 Hz
The frequency of oscillation of the system is given by
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
where
k = 80 N/m is the spring constant
m = 2.90 kg is the mass attached to the spring
Substituting the numbers into the formula, we find
[tex]f=\frac{1}{2\pi}\sqrt{\frac{80 N/m}{2.90 kg}}=0.84 Hz[/tex]
(c) 1.05 m/s
The maximum speed of a spring-mass system is given by
[tex]v=\omega A[/tex]
where
[tex]\omega[/tex] is the angular frequency
A is the amplitude of the motion
For this system, we have
[tex]\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s[/tex]
[tex]A=0.200 m[/tex] (the amplitude corresponds to the maximum displacement, so it is equal to the initial displacement)
Substituting into the formula, we find the maximum speed:
[tex]v=(5.25 rad/s)(0.200 m)=1.05 m/s[/tex]
(d) x = 0
The maximum speed in a simple harmonic motion occurs at the equilibrium position. In fact, the total mechanical energy of the system is equal to the sum of the elastic potential energy (U) and the kinetic energy (K):
[tex]E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2[/tex]
where
k is the spring constant
x is the displacement
m is the mass
v is the speed
The mechanical energy E is constant: this means that when U increases, K decreases, and viceversa. Therefore, the maximum kinetic energy (and so the maximum speed) will occur when the elastic potential energy is minimum (zero), and this occurs when x=0.
(e) 5.51 m/s^2
In a simple harmonic motion, the maximum acceleration is given by
[tex]a=\omega^2 A[/tex]
Using the numbers we calculated in part c):
[tex]\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s[/tex]
[tex]A=0.200 m[/tex]
we find immediately the maximum acceleration:
[tex]a=(5.25 rad/s)^2(0.200 m)=5.51 m/s^2[/tex]
(f) At the position of maximum displacement: [tex]x=\pm 0.200 m[/tex]
According to Newton's second law, the acceleration is directly proportional to the force on the mass:
[tex]a=\frac{F}{m}[/tex]
this means that the acceleration will be maximum when the force is maximum.
However, the force is given by Hooke's law:
[tex]F=kx[/tex]
so, the force is maximum when the displacement x is maximum: so, the maximum acceleration occurs at the position of maximum displacement.
(g) 1.60 J
The total mechanical energy of the system can be found by calculating the kinetic energy of the system at the equilibrium position, where x=0 and so the elastic potential energy U is zero. So we have
[tex]E=K=\frac{1}{2}mv_{max}^2[/tex]
where
m = 2.90 kg is the mass
[tex]v_{max}=1.05 m/s[/tex] is the maximum speed
Solving for E, we find
[tex]E=\frac{1}{2}(2.90 kg)(1.05 m/s)^2=1.60 J[/tex]
(h) 0.99 m/s
When the position is equal to 1/3 of the maximum displacement, we have
[tex]x=\frac{1}{3}(0.200 m)=0.0667 m[/tex]
so the elastic potential energy is
[tex]U=\frac{1}{2}kx^2=\frac{1}{2}(80 N/m)(0.0667 m)^2=0.18 J[/tex]
and since the total energy E = 1.60 J is conserved, the kinetic energy is
[tex]K=E-U=1.60 J-0.18 J=1.42 J[/tex]
And from the relationship between kinetic energy and speed, we can find the speed of the system:
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.42 J)}{2.90 kg}}=0.99 m/s[/tex]
(i) 1.84 m/s^2
When the position is equal to 1/3 of the maximum displacement, we have
[tex]x=\frac{1}{3}(0.200 m)=0.0667 m[/tex]
So the restoring force exerted by the spring on the mass is
[tex]F=kx=(80 N/m)(0.0667 m)=5.34 N[/tex]
And so, we can calculate the acceleration by using Newton's second law:
[tex]a=\frac{F}{m}=\frac{5.34 N}{2.90 kg}=1.84 m/s^2[/tex]
The spring constant of the dense object will be 80N/m.
How to calculate the spring constant?The spring constant will be calculated thus:
k = f/m = 16/0.2 = 80N/m.
The frequency oscillation will be:
= 1/2π × ✓k/✓m
= 1/2π × ✓80/✓2.9
= 0.84 Hz
The maximum speed of the spring mass system will be:
v = [2π(0.84)] × 0.2
= 1.05m/s
The position on the x-axis where the maximum speed occur is at x = 0.
The maximum acceleration will be:
a = w²A
a = 5.25² × 0.2
= 5.51m/s²
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A parallel-plate capacitor initially has air (K= 1) between the plates. We first fullycharge it by a 12 V battery. After the battery is disconnected, we insert a dielectricbetween the plates and it completely fills the space in between. A voltmeter is placedacross the capacitor and it reads 3.6 V now.(a) Assuming that there is no charge loss during the process, what is the dielectricconstant of this material?(b) What’s the fraction of the stored energy changed by inserting the dielectric?(c) What will the voltmeter read if the dielectric is pulled partway out so that it fillsonly half of the space in between the plates?
(a) 3.33
For a capacitor with dielectric disconnected from the battery, the relationship between the voltage across the capacitor without the dielectric (V) and with the dielectric (V) is given by
[tex]V' = \frac{V}{k}[/tex]
where
k is the dielectric constant of the material
In this problem, we have
[tex]V' = 3.6 V[/tex]
[tex]V=12 V[/tex]
So we can re-arrange the formula to find the dielectric constant:
[tex]k=\frac{V}{V'}=\frac{12 V}{3.6 V}=3.33[/tex]
(b) The energy stored reduces by a factor 3.33
The energy stored in a capacitor is
[tex]U=\frac{1}{2}QV[/tex]
where
Q is the charge stored on the capacitor
V is the voltage across the capacitor
Here we can write the initial energy stored in the capacitor (without dielectric) as
[tex]U=\frac{1}{2}QV[/tex]
while after inserting the dielectric is
[tex]U'=\frac{1}{2}QV' = \frac{1}{2}Q\frac{V}{k}[/tex]
since Q, the charge, has not changed (the capacitor is disconnected, so the charge cannot flow away from the capacitor).
So the ratio between the two energies is
[tex]\frac{U'}{U}=\frac{\frac{1}{2}Q\frac{V}{k}}{\frac{1}{2}QV}=\frac{1}{k}[/tex]
which means
[tex]U' = \frac{U}{k}=\frac{U}{3.33}[/tex]
So, the energy stored has decreased by a factor 3.33.
(c) 5.5 V
Pulling the dielectric only partway so that it fills half of the space between the plates is equivalent to a system of 2 capacitors in parallel, each of them with area A/2 (where A is the original area of the plates of the capacitor), of which one of the two is filled with dielectric while the other one is not.
Calling
[tex]C=\frac{\epsilon_0 A}{d}[/tex] the initial capacitance of the capacitor without dielectric
The capacitance of the part of the capacitor of area A/2 without dielectric is
[tex]C_1 = \frac{\epsilon_0 \frac{A}{2}}{d}= \frac{C}{2}[/tex]
while the capacitance of the part of the capacitor with dielectric is
[tex]C_2 = \frac{k \epsilon_0 \frac{A}{2}}{d}= \frac{kC}{2}[/tex]
The two are in parallel, so their total capacitance is
[tex]C' = C_1 + C_2 = \frac{C}{2}+\frac{kC}{2}=(1+k)\frac{C}{2}=(1+3.33)\frac{C}{2}=2.17 C[/tex]
We also have that
[tex]V=\frac{Q}{C}=12 V[/tex] this is the initial voltage
So the final voltage will be
[tex]V' = \frac{Q}{C'}=\frac{Q}{2.17 C}=\frac{1}{2.17}V=\frac{12 V}{2.17}=5.5 V[/tex]
Final answer:
The dielectric constant of the material is 3.3 based on the voltage drop after insertion between the capacitor's plates from 12 V to 3.6 V. The insertion of the dielectric changes the stored energy to 9% of the original. With half the space filled by the dielectric, the voltmeter would read approximately 5.6 V.
Explanation:
The question concerns the concept of capacitors in physics, specifically the effects of inserting a dielectric material between the plates of a parallel-plate capacitor. Assuming that the battery has been disconnected and there is no charge loss during the process, let's address the parts of the question step by step.
(a) Dielectric constant of the material
The dielectric constant (K) of a material can be determined by the ratio of the initial voltage (Vo) to the voltage (V) after the dielectric has been inserted. Given that the initial voltage is 12 V and it drops to 3.6 V, the dielectric constant is K = Vo / V = 12 / 3.6 = 3.3. Therefore, the dielectric constant of this material is 3.3.
(b) Fraction of the stored energy changed
The stored energy in a capacitor is proportional to the square of the voltage across it. When the voltage drops due to the insertion of the dielectric, the energy stored changes by a factor of (V / Vo)2. Hence, the fraction of the energy change is (3.6 / 12)2 = 0.09, or 9% of the original stored energy.
(c) Voltmeter reading with half space filled by the dielectric
When the dielectric only fills half of the space between the plates, the effective dielectric constant becomes a weighted average of the dielectric and air (K=1). Assuming the dielectric constant of the inserted material is 3.3, and it fills half the space, the effective dielectric constant is (3.3 + 1) / 2 = 2.15. Thus, the new voltage is Vnew = Vo / Keff = 12 / 2.15 = 5.58 V. Therefore, the voltmeter will read approximately 5.6 V when the dielectric fills only half of the space between the plates.
At sunset, red light travels horizontally through the doorway in the western wall of your beach cabin, and you observe the light on the eastern wall. Assume that the light has a wavelength of 700 nm, that the door is 1.0 m wide, and that the distance from the door to the far wall of your cabin in 14 m .Part ADetermine the distance between the central bright fringe and a first-order dark fringe of the interference pattern created by the doorway "slit".
Answer:
[tex]9.8\cdot 10^{-6}m[/tex]
Explanation:
For light passing through a single slit, the position of the nth-minimum from the central bright fringe in the diffraction pattern is given by
[tex]y=\frac{n \lambda D}{d}[/tex]
where
[tex]\lambda[/tex] is the wavelength
D is the distance of the screen from the slit
d is the width of the slit
In this problem, we have
[tex]\lambda=700 nm = 7.00\cdot 10^{-7}m[/tex] is the wavelength of the red light
D = 14 m is the distance of the screen from the doorway
d = 1.0 m is the width of the doorway
Substituting n=1 into the equation, we find the distance between the central bright fringe and the first-order dark fringe (the first minimum):
[tex]y=\frac{(1)(7.00\cdot 10^{-7} m)(14 m)}{1.0 m}=9.8\cdot 10^{-6}m[/tex]
List the main types of electromagnetic waves in order of increasing frequency. radio waves, microwaves, infrared, visible light, ultraviolet light, X-rays, gamma rays microwaves, radio waves, infrared, visible light, ultraviolet light, X-rays, gamma rays radio waves, microwaves, infrared, ultraviolet light, visible light, X-rays, gamma rays radio waves, microwaves, visible light, infrared, ultraviolet light, X-rays, gamma rays radio waves, microwaves, infrared, visible light, ultraviolet light, gamma rays, X-rays
Answer:
radio waves, microwaves, infrared, visible light, ultraviolet light, X-rays, gamma rays
Explanation:
Electromagnetic waves are oscillations of electric and magnetic fields in a direction perpendicular to the direction of motion of the wave (transverse waves). They are classified into 7 different types, according to their frequencies.
From lowest to highest frequency, we have:
Radio waves [tex]<10^9 Hz[/tex]
Microwaves [tex]10^9 Hz - 4\cdot 10^{13}Hz[/tex]
Infrared [tex]4\cdot 10^{13} - 4\cdot 10^{14} Hz[/tex]
Visible light [tex]4\cdot 10^{14} - 8\cdot 10^{14}Hz[/tex]
Ultraviolet [tex]8\cdot 10^{14} - 2.4\cdot 10^{16} Hz[/tex]
X-rays [tex]2.4\cdot 10^{16} -5 \cdot 10^{19}Hz[/tex]
Gamma rays [tex]>5\cdot 10^{19} Hz[/tex]
The speed of electromagnetic waves (which include visible light, radio, and x rays) in vacuum is 3.0 x 108 m/s. Wavelengths of visible light waves range from about 400 nm in the violet to about 700 nm in the red. What are the (a) minimum and (b) maximum frequencies of these waves? The range of frequencies for shortwave radio (for example, FM radio and VHF television) is 1.5 to 300 MHz. What are the (c) minimum and (d) maximum wavelengths of these waves? X ray wavelengths range from about 5.0 nm to about 1.0 x 10-2 nm. What are the (e) minimum and (f) maximum frequencies of x rays?
(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.
A train is moving toward the station at a speed of 25 m/s. Its horn emits a sound of frequency 600 Hz. (a) Calculate the frequency detected by a person standing still at the station. (Use 345 m/s for the speed of sound.) (b) As the train moves away, still blowing its horn at the same frequency as before, the observer hears a frequency of 567 Hz. Calculate the new speed of the train.
(a) 646.9 Hz
The formula for the Doppler effect is:
[tex]f'=\frac{v+v_o}{v-v_s}f[/tex]
where
f = 600 Hz is the real frequency of the sound
f' is the apparent frequency
v = 345 m/s is the speed of sound
[tex]v_o = 0[/tex] is the velocity of the observer (zero since it is stationary at the station)
[tex]v_s = +25 m/s[/tex] is the velocity of the source (the train), moving toward the observer
Substituting into the formula,
[tex]f'=\frac{345 m/s+0}{345 m/s-25 m/s}(600 Hz)=646.9 Hz[/tex]
(b) 20.1 m/s
In this case, we have
f = 600 Hz is the real frequency
f' = 567 Hz is the apparent frequency
Assuming the observer is still at rest,
[tex]v_o = 0[/tex]
so we can re-arrange the Doppler formula to find [tex]v_s[/tex], the new velocity of the train:
[tex]f'=\frac{v}{v-v_s}f\\\frac{f}{f'}=\frac{v-v_s}{v}\\\frac{f}{f'}v=v-v_s\\v_s = (1-\frac{f}{f'})v=(1-\frac{600 Hz}{567 Hz})(345 m/s)=-20.1 m/s[/tex]
and the negative sign means the train is moving away from the observer at the station.
The observed frequency of the train's horn is 600 Hz for a stationary observer. As the train moves away, its speed is calculated to be approximately 14.81 m/s to yield an observed frequency of 567 Hz.
Explanation:This question revolves around the principle of Doppler Effect in Physics. The Doppler Effect explains how observed frequency shifts depending on the relative motion of the source (in this case, a train) and the observer (a person at the station).
(a) As the train approaches the person, the frequency detected will be higher than the original. This can be calculated using the formula: f' = f((v + vo) / v), where f' is the observed frequency, f is the source frequency, v is the speed of sound, and vo is the observer's velocity. Since the observer is stationary, vo = 0. f' = 600 Hz * ((345 m/s + 0) / 345 m/s) = 600 Hz
(b) As the train moves away from the person, the observed frequency decreases. Use the formula: f' = f * (v / (v + vs)), where vs is the source's velocity. Solving for vs, we get vs = v - (v * f / f'), equals to 345 m/s - (345 m/s * 600 Hz / 567 Hz) = 14.81 m/s.
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Which statement about electric and magnetic fields is true
A. electric fields must leave a positive charge and end on a negative charge
B. electric fields and magnetic fields start at a positive and end at a negative.
C. Magnetic fields must start at the north pole of a magnet and end at the south pole of a magnet.
D. electric fields loop from south to north
Statement C about electric and magnetic fields is true. Option C is correct.
What is a magnetic field?It is the type of field where the magnetic force is obtained. With the help of a magnetic field. The magnetic force is obtained, it is the field felt around a moving electric charge.
There are two poles of the magnet;
1. North Pole.
2. South Pole.
It has two poles, one north, and one south, and when hanging freely, the magnet aligns itself such that the north pole faces the earth's magnetic north pole.
They point away from the north pole and towards the south pole, is the statement correctly describing magnetic field lines around a magnet.
Statement about electric and magnetic fields is true is;
Magnetic fields must start at the north pole of a magnet and end at the south pole of a magnet.
Hence, option C is correct.
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A sled is being pulled along a horizontal surface by a horizontal force F of magnitude 600 N. Starting from rest, the sled speeds up with acceleration 0.08 m/s^2 for 1 minute. Find the average power P created by force F.
Answer:
1440 W
Explanation:
First of all, we can find the total displacement of the sled, which is given by
[tex]d=\frac{1}{2}at^2[/tex]
where
a = 0.08 m/s^2 is the acceleration
t = 1 min = 60 s is the time
Substituting,
[tex]d=\frac{1}{2}(0.08 m/s^2)(60 s)^2=144 m[/tex]
Now we can find the wotk done on the sled, equal to the product between force and displacement:
[tex]W=Fd=(600 N)(144 m)=86,400 J[/tex]
And finally we can fidn the average power, which is the ratio between the work done and the time taken:
[tex]P=\frac{W}{t}=\frac{86,400 J}{60 s}=1440 W[/tex]
A uniform square crate is released from rest with corner D directly above A and it rotates about A until its corner B impacts the floor and then it rotates about B. The floor is rough enough to prevent slipping and the corner at B does not rebound on impact. Determine, (a) as a function of g and `, the angular velocity of the crate !E2 just before impact. (b) as a function of !E2, the angular velocity of the crate immediately after impact !E3. (c) the fraction of energy lost during the impact. (d) the maximum achieved after the impact. During the impact, assume that the weight force is non-impulsive.
(c) the fraction of energy lost during the impact
Boxing gloves are padded to lessen the force of a blow. (a) Calculate the force exerted by a boxing glove on an opponent’s face, if the glove and face compress 7.50 cm during a blow in which the 7.00-kg arm and glove are brought to rest from an initial speed of 10.0 m/s. (b) Calculate the force exerted by an identical blow in the days when no gloves were used and the knuckles and face would compress only 2.00 cm. (c) Discuss the magnitude of the force with glove on. Does it seem high enough to cause damage even though it is lower than the force with no glove?
(a) -4667 N
First of all, we can calculate the acceleration of the arm and the glove, using the following equation:
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 0 is the final speed
u = 10 m/s is the initial speed
a is the acceleration
d = 7.50 cm = 0.075 m is the distance through which the arm and the glove move before coming to a stop
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-(10.0 m/s)^2}{2(0.075 m)}=-666.7 m/s^2[/tex]
And since the know the mass of the arm+glove:
m = 7.00 kg
We can now calculate the force exerted:
[tex]F=ma=(7.00 kg)(-666.7 N)=-4,667 N[/tex]
(b) -17500 N
We can repeat the problem, but this time the stopping distance is different:
d = 2.00 cm = 0.02 m
So the acceleration is
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-(10.0 m/s)^2}{2(0.02 m)}=-2500 m/s^2[/tex]
and so the force is
[tex]F=ma=(7.00 kg)(-2500 m/s^2)=-17,500 N[/tex]
(c) Yes
The force exerted when the glove is used is
F = 4667 N
We see that this force corresponds approximately to the weight of an object of mass m=476 kg, in fact:
[tex]W=mg=(476 kg)(9.81 m/s^2)=4670 N[/tex]
Which is quite a lot. Therefore, the force even when gloves are used seems enough to cause damage.
To calculate the force exerted by a boxing glove on an opponent's face, we can use the equation for impulse.
Explanation:In order to calculate the force exerted by a boxing glove on an opponent's face, we need to use the equation for impulse:
Impulse = change in momentum = force x time
We know the initial speed of the arm and glove (10.0 m/s), the final speed (0 m/s), and the time it takes for the glove and face to compress (which can be calculated using the distance and the initial speed).
By substituting these values into the equation, we can find the average force exerted by the boxing glove on the opponent's face.
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The difference between the mass of the nucleus and the mass of the nucleons it contains is converted into ? that holds the nucleus together
Photoelectric energy
Binding energy
Answer:
Binding energy
Explanation:
The energy required to bind the nucleus is called binding energy.
Binding energy is equal to the product of mass defect and square of velocity of light.
Mass defect is the difference of mass of neucleons and the mass of atom.
Answer:
Binding Energy
Explanation:
Consider a beam of electrons in a vacuum, passing through a very narrow slit of width 2.00μm. The electrons then head toward an array of detectors a distance 1.068 m away. These detectors indicate a diffraction pattern, with a broad maximum of electron intensity (i.e., the number of electrons received in a certain area over a certain period of time) with minima of electron intensity on either side, spaced 0.527 cm from the center of the pattern. What is the wavelength λ of one of the electrons in this beam? Recall that the location of the first intensity minima in a single slit diffraction pattern for light is y=Lλ/a, where L is the distance to the screen (detector) and a is the width of the slit. The derivation of this formula was based entirely upon the wave nature of light, so by de Broglie's hypothesis it will also apply to the case of electron waves.
Answer:
[tex]9.9\cdot 10^{-9}m[/tex]
Explanation:
In a single-slit diffraction pattern, the location of the first minimum is given by
[tex]y=\frac{L\lambda }{a}[/tex]
where
L is the distance between the slit and the screen
[tex]\lambda[/tex] is the wavelength
a is the width of the slit
In this problem, we have
[tex]y=0.527 cm = 5.27\cdot 10^{-3} m[/tex] (location of first minimum)
L = 1.068 m (distance of the screen)
[tex]a=2.00\mu m=2.00\cdot 10^{-6}m[/tex] (width of the slit)
Solving the equation for [tex]\lambda[/tex], we find the De Broglie wavelength of the electron:
[tex]\lambda = \frac{ya}{L}=\frac{(5.27\cdot 10^{-3} m)(2.00\cdot 10^{-6} m)}{1.068 m}=9.9\cdot 10^{-9}m[/tex]
The question examines the physics concept of diffraction about a beam of electrons passing through a narrow slit. De Broglie's hypothesis applies in this context as particles can exhibit characteristics of waves. The wavelength of the electron wave can be determined using the formula for the location of first-intensity minima, taking into consideration the values provided.
Explanation:The question refers to a phenomenon known as diffraction, which is observed when electrons pass through a narrow slit toward a detector and create a diffraction pattern. The principle in question is the de Broglie hypothesis which maintains that particles can exhibit wave-like behavior. To calculate the wavelength λ of one of the electrons in this beam, we use the location of the first intensity minima formula: y=Lλ/a. In the given scenario, y is the distance from the center of the intensity minima to the broad maximum (0.527 cm), L is the distance to the screen (1.068 m), and a is the width of the slit (2.00μm). Using these values, you can solve for λ, the wavelength of the electron wave.
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Coders play an important role in
Answer:
it is a.health record documentation
Explanation:hope this helps
Answer: Reimbursement and research
Explanation: I got it right on the PF exam
What’s The Answer, To The Question In The Photo
Answer:
The correct answer is the third option: The kinetic energy of the water molecules decreases.
Explanation:
Temperature is, in depth, a statistical value; kind of an average of the particles movement in any physical system (such as a glass filled with water). Kinetic energy, for sure, is the energy resulting from movement (technically depending on mass and velocity of a system; in other words, the faster something moves, the greater its kinetic energy.
Since temperature is related to the total average random movement in a system, and so is the kinetic energy (related to movement through velocity), as the thermometer measures less temperature, that would mean that the particles (in this case: water particles) are moving slowly, so that: the slower something moves, the lower its kinetic energy.
In summary: temperature tells about how fast are moving and colliding the particles within a system, and since it is directly proportional to the amount of movement, it can be related (also directly proportional) to the kinectic energy.
What is EMF?
When are we exposed to EMF?
Does EMF increase or decrease with distance?
What are the primary differences in direct current and alternating current?
Answer:
Electricity is the voltage produced by any electrical energy source such as battery or dynamo. EMF increase with distance.
The direct current is fixed but the alternating current is constantly changing
Which actions are ways to manipulate the magnet and wire loop in order to induce a current? Check all that apply. Move the magnet forward. Move the loop away from the magnet. Place the loop right next to the magnet, both stationary. Move the loop sideways. Move the magnet and loop in the same direction, at the same speed, at the same time. Rotate the loop in place.
Answer:
1, 2, 4, 6
Explanation:
Move the magnet forward.
Move the loop away from the magnet.
Move the loop sideways.
Rotate the loop in place.
Answer:
Explanation:
When a changing magnetic flux linked with the coil, then the induced emf or induced current is developed.
1. Move the magnet forward.
Here, the flux linked with the coil changes, so the induced current is developed.
2. Move the loop away from the magnet.
Here, the flux linked with the coil changes, so the induced current is developed.
3. Place the loop right next to the magnet, both stationary.
Here, the flux linked with the coil does not change, so the induced current is not developed.
4. Move the loop sideways.
Here, the flux linked with the coil changes, so the induced current is developed.
5. Move the magnet and loop in the same direction, at the same speed, at the same time.
Here, the flux linked with the coil does not change, so the induced current is not developed.
6. Rotate the loop in place.
Here, the flux linked with the coil changes, so the induced current is developed.
Use Snell's Law to solve the following:
Answer:
1.171
Explanation:
if n₁sinΘ₁=n₂sinΘ₂, then n₂=n₁sinΘ₁ / sinΘ₂;
[tex]n_2=\frac{1.5*sin45}{sin65}=\frac{1.5*0.707}{0.906} =1.1705[/tex]
The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassium 2.2 Sodium 2.3 Lithium 2.5 Calcium 3.2 Copper 4.5 Silver 4.7 Platinum 5.6 Using these data, answer the following questions about the photoelectric effect. Part A Light with a wavelength of 190 nm is incident on a metal surface. The most energetic electrons emitted from the surface are measured to have 4.0 eV of kinetic energy. Which of the metals in the table is the surface most likely to be made of? View Available Hint(s) Submit Part B Of the eight metals listed in the table, how many will eject electrons when a green laser (λg=510nm) is shined on them? View Available Hint(s) Submit Part C Light with some unknown wavelength is incident on a piece of copper. The most energetic electrons emitted from the copper have 2.7 eV of kinetic energy. If the copper is replaced with a piece of sodium, what will be the maximum possible kinetic energy K of the electrons emitted from this new surface? Enter your answer numerically in electron volts to two significant figures. View Available Hint(s)
A. Lithium
The equation for the photoelectric effect is:
[tex]E=\phi + K[/tex]
where
[tex]E=\frac{hc}{\lambda}[/tex] is the energy of the incident light, with h being the Planck constant, c being the speed of light, and [tex]\lambda[/tex] being the wavelength
[tex]\phi[/tex] is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)
K is the maximum kinetic energy of the photoelectron
In this problem, we have
[tex]\lambda=190 nm=1.9\cdot 10^{-7}m[/tex], so the energy of the incident light is
[tex]E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J[/tex]
Converting in electronvolts,
[tex]E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV[/tex]
Since the electrons are emitted from the surface with a maximum kinetic energy of
K = 4.0 eV
The work function of this metal is
[tex]\phi = E-K=6.5 eV-4.0 eV=2.5 eV[/tex]
So, the metal is Lithium.
B. cesium, potassium, sodium
The wavelength of green light is
[tex]\lambda=510 nm=5.1\cdot 10^{-7} m[/tex]
So its energy is
[tex]E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J[/tex]
Converting in electronvolts,
[tex]E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV[/tex]
So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:
Cesium
Potassium
Sodium
C. 4.9 eV
In this case, we have
- Copper work function: [tex]\phi = 4.5 eV[/tex]
- Maximum kinetic energy of the emitted electrons: K = 2.7 eV
So, the energy of the incident light is
[tex]E=\phi+K=4.5 eV+2.7 eV=7.2 eV[/tex]
Then the copper is replaced with sodium, which has work function of
[tex]\phi = 2.3 eV[/tex]
So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be
[tex]K=E-\phi = 7.2 eV-2.3 eV=4.9 eV[/tex]
The surface is most likely to be made of Lithium. Four metals (Cesium, Potassium, Sodium, and Lithium) will eject electrons when a green laser is shone on them. The maximum kinetic energy of the electrons emitted from Sodium will be approximately 4.9 eV.
Explanation:Part A: Let's calculate the energy of the incident photon using Planck's equation E = h * c / λ , where h is Planck's constant (6.626 * 10^-34 J.s), c is the speed of light (3 * 10^8 m/s) and λ is the wavelength of the incident light (convert 190 nm to 1.9 * 10^-7 m). The energy of the photon (Ep) is approximately 6.6 eV. The given kinetic energy (Ek) of the most energetic electrons is 4.0 eV. According to the photoelectric effect, Ep = Work Function + Ek. So, the work function of the metal surface (Wf) is Ep - Ek, which gives us a work function of 2.6 eV. The closest value to this in the table is that of Lithium with 2.5 eV, so the surface is likely made of Lithium.
Part B: To find out the number of metals that will eject electrons when a green laser of wavelength 510 nm is shone on them, we first need to convert this wavelength into energy using the same formula as in Part A. The energy hence calculated is approximately 2.4 eV. As per the photoelectric effect, a metal can only eject electrons when its work function is less than or equal to the energy of the incident light. Therefore, only metals with work functions less than or equal to 2.4 eV will eject electrons. Looking at the table, we see that this covers Cesium, Potassium, Sodium, and Lithium. So, 4 metals will eject electrons.
Part C: Given the kinetic energy of electrons emitted from copper is 2.7 eV, and the work function of copper is 4.5 eV. According to the photoelectric effect, the energy of the incident radiation should be the sum of these two, which yields 7.2 eV. Now, if we replace the copper sample with sodium (having a work function of 2.3 eV), then the maximum kinetic energy of the emitted electrons from the sodium sample (Ek) will be the energy of the incident light minus the work function of Sodium i.e., 7.2 eV - 2.3 eV which is approximately 4.9 eV.
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An electron in a cathode-ray beam passes between 2.5-cm-long parallel-plate electrodes that are 5.6 mm apart. A 2.5 mT , 2.5-cm-wide magnetic field is perpendicular to the electric field between the plates. The electron passes through the electrodes without being deflected if the potential difference between the plates is 600 V . Part A What is the electron's speed? Express your answer to two significant figures and include the appropriate units. v v = nothingnothing SubmitRequest Answer Part B If the potential difference between the plates is set to zero, what is the electron's radius of curvature in the magnetic field? Express your answer to two significant figures and include the appropriate units. r r = nothingnothing
A) [tex]4.3\cdot 10^7 m/s[/tex]
For an electron moving in a region with both electric and magnetic field, the electron will move undeflected if the electric force on the electron is equal to the magnetic force:
[tex]qE= qvB[/tex]
which means that the speed of the electron will be
[tex]v=\frac{E}{B}[/tex]
where
E is the magnitude of the electric field
B is the magnitude of the magnetic field
In this problem,
[tex]B=2.5 mT=0.0025 T[/tex] is the intensity of the magnetic field
The electric field can be found as
[tex]E=\frac{V}{d}[/tex]
where
V = 600 V is the potential difference between the electrodes
[tex]d=5.6 mm=0.0056 m[/tex] is the distance between the electrodes
Substituting,
[tex]E=\frac{600 V}{0.0056 m}=1.07\cdot 10^5 V/m[/tex]
So the electron's speed is
[tex]v=\frac{1.07\cdot 10^5 V/m}{0.0025 T}=4.3\cdot 10^7 m/s[/tex]
B) [tex]9.8\cdot 10^{-2} m[/tex]
The radius of curvature of an electron in a magnetic field can be found by equalizing the centripetal force to the magnetic force:
[tex]m\frac{v^2}{r}=qvB[/tex]
where
m is the electron mass
v is the speed
r is the radius of curvature
q is the charge of the electron
Solving for r, we find
[tex]r=\frac{mv}{qB}=\frac{(9.11\cdot 10^{-31} kg)(4.3\cdot 10^7 m/s)}{(1.6\cdot 10^{-19} C)(0.0025 T)}=9.8\cdot 10^{-2} m[/tex]
A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric field, of amplitude 310 V/m, oscillates parallel to the y axis. What are the (a) frequency, (b) angular frequency, and (c) angular wave number of the wave? (d) What is the amplitude of the magnetic field component? (e) Parallel to which axis does the magnetic field oscillate? (f) What is the time-averaged rate of energy flow associated with this wave? The wave uniformly illuminates a surface of area 1.8 m2. If the surface totally absorbs the wave, what are (g) the rate at which momentum is transferred to the surface and (h) the radiation pressure?
(a) [tex]7.32\cdot 10^7 Hz[/tex]
The frequency of an electromagnetic waves is given by:
[tex]f=\frac{c}{\lambda}[/tex]
where
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
[tex]\lambda=4.1 m[/tex] is the wavelength of the wave in the problem
Substituting into the equation, we find
[tex]f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz[/tex]
(b) [tex]4.60\cdot 10^8 rad/s[/tex]
The angular frequency of a wave is given by
[tex]\omega = 2\pi f[/tex]
where
f is the frequency
For this wave,
[tex]f=7.32\cdot 10^7 Hz[/tex]
So the angular frequency is
[tex]\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s[/tex]
(c) [tex]1.53 m^{-1}[/tex]
The angular wave number of a wave is given by
[tex]k=\frac{2\pi}{\lambda}[/tex]
where
[tex]\lambda[/tex] is the wavelength of the wave
For this wave, we have
[tex]\lambda=4.1 m[/tex]
so the angular wave number is
[tex]k=\frac{2\pi}{4.1 m}=1.53 m^{-1}[/tex]
(d) [tex]1.03\cdot 10^{-6}T[/tex]
For an electromagnetic wave,
[tex]E=cB[/tex]
where
E is the magnitude of the electric field component
c is the speed of light
B is the magnitude of the magnetic field component
For this wave,
E = 310 V/m
So we can re-arrange the equation to find B:
[tex]B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T[/tex]
(e) z-axis
In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:
- direction of propagation of the wave --> positive x axis
- direction of oscillation of electric field --> y axis
- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis
(f) 127.5 W/m^2
The time-averaged rate of energy flow of an electromagnetic wave is given by:
[tex]I=\frac{E^2}{2\mu_0 c}[/tex]
where we have
E = 310 V/m is the amplitude of the electric field
[tex]\mu_0[/tex] is the vacuum permeability
c is the speed of light
Substituting into the formula,
[tex]I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2[/tex]
(g) [tex]1.53\cdot 10^{-8} kg m/s[/tex]
For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by
[tex]\frac{dp}{dt}=\frac{<S>A}{c}[/tex]
where the <S> is the magnitude of the Poynting vector, given by
[tex]<S>=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2[/tex]
and where the surface is
A = 1.8 m^2
Substituting, we find
[tex]\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s[/tex]
(h) [tex]8.47\cdot 10^{-7} N/m^2[/tex]
For a surface that totally absorbs the wave, the radiation pressure is given by
[tex]p=\frac{<S>}{c}[/tex]
where we have
[tex]<S>=254.2 W/m^2[/tex]
[tex]c=3\cdot 10^8 m/s[/tex]
Substituting, we find
[tex]p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2[/tex]
What's true about the elliptical path that the planets follow around the sun? A. A line can be drawn from the planet to the sun that sweeps out equal areas in equal times. B. A line can be drawn from the planet to the sun that follows the same curve as the ellipse. C. A scalar can be measured from the angle that the planet travels relative to the sun's orbit. D. A vector can be drawn from the center of one planet to the center of an adjacent planet.
Answer:
A. A line can be drawn from the planet to the sun that sweeps out equal areas in equal times
Explanation:
This is exactly what Kepler's second law of planetary motion states:
"the segment joining the sun with the center of each planet sweeps out equal areas in equal time"
This law basically tells how the speed of a planet orbiting the sun changes during its revolution. In fact, we have that:
- when a planet is closer to the Sun, it will orbit faster
- when a planet is farther from the Sun, it will orbit slower
Which of the following is the least reliable source of background information for a scientific project?
general Internet site
government Internet site
library reserve section
scientific journal
Answer:
general internet site
Explanation:
Answer:
General internet site
Explanation:
A technician wearing a brass bracelet enclosing area 0.00500 m2 places her hand in a solenoid whose magnetic field is 2.70 T directed perpendicular to the plane of the bracelet. The electrical resistance around the circumference of the bracelet is 0.0200 . An unexpected power failure causes the field to drop to 0.81 T in a time of 19.0 ms. (a) Find the current induced in the bracelet. A (b) Find the power (????) delivered to the bracelet. (Note: As this problem implies, you should not wear any metal objects when working in regions of strong magnetic fields.) W
(a) 25 A
The induced emf in the circuit is given by
[tex]\epsilon = -\frac{\Delta \Phi}{\Delta t}[/tex]
where
[tex]\Delta Phi[/tex] is the variation of magnetic flux through the bracelet
[tex]\Delta t = 19.0 ms =0.019 s[/tex] is the time interval
The variation of magnetic flux is
[tex]\Delta \Phi = A \Delta B[/tex]
where
[tex]A=0.005 m^2[/tex] is the area enclosed by the bracelet
[tex]\Delta B=0.81 T-2.70 T=-1.89 T[/tex] is the change of magnetic field strength
So we find
[tex]\Delta \Phi = (0.005 m^2)(-1.89 T)=-9.45\cdot 10^{-3} Wb[/tex]
And so the induced emf is
[tex]\epsilon = -\frac{-9.45\cdot 10^{-3} Wb}{0.019 s}=0.50 V[/tex]
Since the resistance of the bracelet is
[tex]R=0.02\Omega[/tex]
the induced current is
[tex]I=\frac{V}{R}=\frac{0.50 V}{0.02 \Omega}=25 A[/tex]
(b) 12.5 W
The power delivered to the bracelet is given by
[tex]P=V I[/tex]
where we have
I = 25 A is the current
V = 0.50 V is the emf induced in the bracelet
Substituting numbers, we find
[tex]P=(0.50 V)(25 A)=12.5 W[/tex]
The problem involves using Faraday's Law to compute for the induced current and power in the bracelet due to a change in the magnetic field strength of the solenoid it is placed in.
Explanation:This involves calculating the induced current (part a) and power (part b) in the bracelet due to a change in the magnetic field. This can be solved using Faraday's Law which states that the induced electromotive force (EMF) in a circuit is equal to the negative rate of change of magnetic flux. For part a, we compute for the EMF using the formula ΔΦ/Δt = (2.70 T - 0.81 T) * 0.00500 m²/0.019 s which can then be used to calculate the induced current by dividing the EMF by the resistance of the bracelet. Part b involves finding the delivered power using the formula P=I²R, where I is the induced current and R is the resistance.
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17.Explain the different ways that an object can become electrically charged.
18.What rules exist for how charged objects act around each other.
19.Compare and contrast the electrical force and the gravitational force.
17.
There are three different methods for charging objects:
- Friction: in friction, two objects are rubbed against each other. As a result, electrons can be passed from one object to the other, so one object will gain a net negative charge while the other object will gain a net positive charge due to the lack of electrons.
- Conduction: this occurs when two conductive objects are put in contact with each other, and charges (electrons, usually) are transferred from one object to the other one.
- Induction: this occurs when two objects are brought closer to each other, but not in contact. If one of the two objects has a net charge (different from zero) on its surface, then it will induce a movement of charges in the second object: in particular, in the second object, charges of the opposite polarity will be attracted towards the first object, while charges of same polarity will be repelled further away.
18.
Charged objects produce around themselves an electric field. The strenght of the electric field is given by (assuming the charged objects are spherical)
[tex]E=k\frac{q}{r^2}[/tex]
where k is the Coulomb's constant, q is the magnitude of the charge and r the distance from the centre of the charge. As we see, the strength of the field is inversely proportional to the square of the distance.
Also, the direction of the field is determined by the sign of the charge:
- if the charge is positive, the electric field points away from the charge (this means that other positive charges in the field will be repelled away)
- if the charge is negative, the electric field points towards the charge (this means that other positive charges in the field will be attracted towards it)
19.
Electrical force 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, and r their separation.
Gravitational force is given by:
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r their separation.
Similarities between the two forces:
- Both are inversely proportional to the square of the distance between the two objects, r
- Both are non-contact forces (the two objects can experience the forces even if they are not in contact)
- Both forces have infinite range
Differencies between the two forces:
- The electric force can be either attractive or repulsive, while the gravitational force is attractive only
- The electric force is much stronger than the gravitational force, due to the much larger value of the Coulomb's constant k compared to the gravitational constant G
In our solar system, the most likely planet (other than Earth) to have life on it is currently thought to be
Select one:
a. Saturn
b. Jupiter
c. Mars
d. Venus
e. Mercury
Answer: Mars
Explanation:
Mars is a planet similar to the earth which the rest can't sustain life either because it's too hot or cold or too much gas maybe even toxic acid.
Final answer:
Mars is the most likely planet other than Earth to have life in our solar system, based on past habitable conditions. The neighboring terrestrial planets, Venus and Mars, have evolved differently, with Earth being the most habitable. Gas giants like Jupiter and Saturn are not considered habitable.
Explanation:
Mars is currently thought to be the most likely planet, after Earth, to have life. Despite being colder and drier than Earth, Mars has shown evidence of habitable conditions in the past, making it a prime candidate for potential life forms.
Venus and Mars are the neighboring terrestrial planets diverged significantly in their evolution, with Earth being the most hospitable planet in the solar system. The outer gas giant planets like Jupiter and Saturn are almost certainly not habitable for life as we know it.
An electron, traveling at a speed of 5.90 × 10 6 5.90×106 m/s, strikes the target of an X-ray tube. Upon impact, the electron decelerates to three-quarters of its original speed, with an X-ray photon being emitted in the process. What is the wavelength of the photon?
Answer:
[tex]2.84\cdot 10^{-8} m[/tex]
Explanation:
Due to the law of conservation of energy, the energy of the emitted X-ray photon is equal to the energy lost by the electron.
The initial kinetic energy of the electron is:
[tex]K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(9.11\cdot 10^{-31}kg)(5.90\cdot 10^6 m/s)^2=1.59\cdot 10^{-17}J[/tex]
The electrons decelerates to 3/4 of its speed, so the new speed is
[tex]v_f = \frac{3}{4}v_i = \frac{3}{4}(5.90\cdot 10^6 m/s)=4.425\cdot 10^6 m/s[/tex]
So the final kinetic energy is
[tex]K_f = \frac{1}{2}mv_f^2=\frac{1}{2}(9.11 \cdot 10^{-31} kg)(4.425\cdot 10^6 m/s)^2=8.9\cdot 10^{-18} J[/tex]
So, the energy lost by the electron, which is equal to the energy of the emitted photon, is
[tex]E=K_i - K_f =1.59\cdot 10^{-17} J-8.9\cdot 10^{-18} J=7\cdot 10^{-18} J[/tex]
The wavelength of the photon is related to its energy by
[tex]\lambda=\frac{hc}{E}[/tex]
where h is the Planck constant and c the speed of light. Substituting E, we find
[tex]\lambda=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{7\cdot 10^{-18} J}=2.84\cdot 10^{-8} m[/tex]
To determine the wavelength of the photon emitted when an electron decelerates in an X-ray tube, we can use Planck's equation and the information given about the electron's speed.
Explanation:When an electron decelerates, it emits electromagnetic waves known as X-rays. The wavelength of the X-ray photon can be determined using Planck's equation, E = hv, where E is energy, h is Planck's constant, and v is frequency. Given that the electron's velocity decreases to three-quarters of its original speed and the initial speed is 5.90x10^6 m/s, we can calculate the velocity of the electron after deceleration. Using this value, we can then calculate the wavelength of the photon emitted. Therefore, using the provided information, we can find the wavelength of the photon.
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Guiana dolphins are one of the few mammals able to detect electric fields. In a test of sensitivity, a dolphin was exposed to the electric field of charged electrodes. The electric field was measured by detecting the potential difference between two electrodes located 1.0 cm apart along the field lines. The dolphin could reliably detect a field that produced a potential difference of 0.50 mV. What is the corresponding electric field strength?
Answer:
0.05 V/m
Explanation:
For a uniform electric field, electric field strength and potential difference are related by
[tex]E=\frac{\Delta V}{d}[/tex]
where
E is the electric field strength
[tex]\Delta V[/tex] is the potential difference
d is the distance between the two points
Here we have
[tex]\Delta V= 0.50 mV=5\cdot 10^{-4}V[/tex]
[tex]d=1.0 cm=0.01 m[/tex]
So, the electric field strength is
[tex]E=\frac{5\cdot 10^{-4} V}{0.01 m}=0.05 V/m[/tex]
If a hypothesis is falsifiable, _____.
it is wrong
it might be wrong
it must be possible to prove it wrong
it is correct
Answer:
it must be possible to prove it wrong
Explanation:
Answer:
it must be possible to prove it wrong
Explanation:
Falsibiable is the word used to make the refernce that it is possible to prove something wrong or to prove that somthing is wrong with a statement ot hypothesis, since the hypothsesis is falsifiable it must be possible to prove it wrong, that is the correct answer.
A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing experiments, one of their tasks is to demonstrate the concept of the escape speed by throwing rocks straight up at various initial speeds. With what minimum initial speed ????esc will the rocks need to be thrown in order for them never to "fall" back to the asteroid? Assume that the asteroid is approximately spherical, with an average density ????=4.49×106 g/m3 and volume ????=3.32×1012 m3 . Recall that the universal gravitational constant is ????=6.67×10−11 N·m2/kg2 .
Answer:
463.4 m/s
Explanation:
The escape velocity on the surface of a planet/asteroid is given by
[tex]v=\sqrt{\frac{2GM}{R}}[/tex] (1)
where
G is the gravitational constant
M is the mass of the planet/asteroid
R is the radius of the planet/asteroid
For the asteroid in this problem, we know
[tex]\rho=4.49\cdot 10^6 g/m^3[/tex] is the density
[tex]V=3.32\cdot 10^{12} m^3[/tex] is the volume
So we can find its mass:
[tex]M=\frac{\rho}{V}=(4.49\cdot 10^6 g/m^3)(3.32\cdot 10^{12}m^3)=1.49\cdot 10^{19} kg[/tex]
Also, the asteroid is approximately spherical, so its volume is given by
[tex]V=\frac{4}{3}\pi R^3[/tex]
where R is the radius. Solving the formula for R, we find its radius:
[tex]R=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{\frac{3(3.32\cdot 10^{12}m^3)}{4\pi}}=9256 m[/tex]
So now we can use eq.(1) to find the escape velocity:
[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.49\cdot 10^{19}kg)}{9256 m}}=463.4 m/s[/tex]
Imagine two parallel, (infinite) planar sheets of perfectly black material, maintained at fixed temperatures TH and TL, respectively, and separated by some finite distance D. Suspended in between and parallel to these sheets are n additional, separate, parallel sheets of material, all with constant reflectivity r. (By separate, we mean that none of the sheets are in actual mechanical contact). When the entire system reaches a steady state, find (a) the net flux density (energy flow per unit time per unit area) of thermal radiation between the original two sheets. (b) How does this compare to the radiation flux that would be present between the two black sheets in the absence of the n intervening sheets? (c) To provide the best insulation, should these intermediate sheets be made of very shiny or very black material, or something in between?
The TIL will support your answer I believe if you just think about of the provide the inslutoin which that mean money is gong fast to the enconmomy