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?

(c) The baseball's wavelength is too small. Option C is correct.

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 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 :)

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.

The period of a particle undergoing simple harmonic motion is defined as the time taken for the particle to complete one complete oscillation.

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:

The 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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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) [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;

The 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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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.

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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Answer:

gamma Ray

Explanation:

Stable or unstable nucleus will have high energy in form of nuclear energy stored in it.

When this energy exist in the unstable state of the nucleus then in order to attain the stability the nucleus will will go to lower energy state.

This energy change is of higher order due to which energy released is of large order and high frequency electromagnetic waves.

so it is given as

[tex]E_2 - E_1 = h\nu[/tex]

since this energy is of higher range so it is given as gamma rays.

so correct answer will be

Gamma Rays

Answer:

Gamma Ray

Explanation:

Gamma Ray is produced furring nuclear decay which occurs in the nucleus of the atom.

Other radiations are produced from outside the nucleus.

Gamma Ray has the shortest wavelength and the highest penetrating power and ionises matter.

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.

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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(a) 3000 V

For two parallel conducting plates, the potential difference between the plates is given by:

[tex]\Delta V=Ed[/tex]

where

E is the magnitude of the electric field

d is the separation between the plates

Here we have:

[tex]E=7.50\cdot 10^4 V/m[/tex] is the electric field

d = 4.00 cm = 0.04 m is the distance between the plates

Substituting,

[tex]\Delta V=(7.50\cdot 10^4 V/m)(0.04 m)=3000 V[/tex]

(b) 750 V

The potential difference between the two plates A and B is

[tex]\Delta V = V_B - V_A = 3000 V[/tex]

Let's take plate A as the plate at 0 volts:

[tex]V_A = 0 V[/tex]

The potential increases linearly going from plate A (0 V) to plate B (3000 V).

So, if the potential difference between A and B, separated by 4 cm, is 3000 V, then the potential difference between A and a point located at 1 cm from A is given by the proportion:

[tex]3000 V : 4 cm = V(1 cm) : 1 cm[/tex]

and solving for V(1 cm) we find:

[tex]V(1 cm)=\frac{(3000 V)(1 cm)}{4 cm}=750 V[/tex]

Potential difference is the difference in electrical potential. when the lowest potential is taken to be at zero volts then the potential 1.00 cm from that plate is 750 V.

A potential difference is defined as the difference in the electrical potential between two points.

Given to us

Distance between the plate, d = 4 cm = 0.04 m

Electric field, E = 7.50×104 V/m .

A.) We know that for two parallel conducting plates, the potential difference between the plate is given as,

[tex]\triangle V = Ed[/tex]

E = Magnitude of the electric field

d = distance between the plates

Substitute the values,

[tex]E = 7.5 \times 10^4 \times 0.04\\\\E = 3000\rm\ V[/tex]

B.) We already know the potential difference between the two plates, therefore,

The potential difference between the two plates,

[tex]\triangle V= V_A - V_B[/tex]

Since, one of the plates is having zero potential, therefore,

[tex]\triangle V= V_A = 3000\rm\ V[/tex]

We know that the potential difference between the two plates is 300 which are 4 cm apart, therefore,

[tex]\dfrac{\triangle V}{d} = \dfrac{V}{1}\\\\V = 750 \rm\ V[/tex]

Hence, when the lowest potential is taken to be at zero volts then the potential 1.00 cm from that plate is 750 V.

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(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.

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]

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.

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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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.

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]

(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

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]

Answer:

undergoes a transition to a quantum state of lower energy

Explanation:

When electrons in an atom move to another quantum state, they emit/absorb a photon according to the following:

- If the electron is moving to a higher energy state, it absorbs a photon (because it needs energy to move to a higher energy level, so it must absorb the energy of the photon)

- if the electron is moving to a lower energy state, it emits a photon (because it releases the excess energy)

In particular, the energy of the absorbed/emitted photon is exactly equal to the difference in energy between the two levels of the electron transition:

[tex]E=|E_1 - E_2|[/tex]

An atom emits a photon when one of its electrons undergoes a transition to a quantum state of lower energy. Hence, the correct option is 3.

This happens when an electron in an excited energy state returns to a lower energy state or the ground state. The energy difference between these states is released as a photon, which corresponds to the energy of the gap between the two levels.

To summarize, the emission of a photon occurs due to:

In contrast, if an electron moves from a lower energy level to a higher one, it must absorb a photon, not emit it.

(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]

Answer:

speed =distance/time

Formula

s = d/t

s = speed

d = distance traveled

t = time elapsed

Thanks, Hope this helps.

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

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.

Answer:

4.42 s

Explanation:

The frequency of the oscillation is given by the ratio between the number of complete oscillations and the time taken:

[tex]f=\frac{N}{t}[/tex]

where for this glider, we have

N = 7.00

t = 31.0 s

Substituting, we find

[tex]f=\frac{7.00}{31.0 s}=0.226 Hz[/tex]

Now we now that the period of oscillation is the reciprocal of the frequency:

[tex]T=\frac{1}{f}[/tex]

So, substituting f = 0.226 Hz, we find:

[tex]T=\frac{1}{0.226 Hz}=4.42 s[/tex]

Final answer:

The period of oscillations for the air-track glider is found by dividing the total time of 31.0 seconds by the number of oscillations, which is 7.00, resulting in a period of approximately 4.43 seconds.

Explanation:

The question involves finding the period of oscillations for an air-track glider attached to a spring. To calculate the period, we use the formula for the period T, which is T = total time / number of oscillations. Given that the glider completes 7.00 oscillations in 31.0 seconds, we can calculate the period by dividing the total time by the number of oscillations.

The calculation would be as follows: T = 31.0 s / 7.00 oscillations, which equals approximately 4.43 seconds per oscillation.

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.

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:

In 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.

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:

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".

56 will be the answer all you do is divide 104 to 26 then divide 2.6

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°

sin(50) = vertical / 4

vertical = sin(50)(4)

            = 3.06

            ≈ 3.1

  cosΔ = adj / hypo

cos(50) = horizontal / 4

horizontal = cos(50)(4)

                = 2.57

                ≈ 2.6

(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.

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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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 :/