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Damm [24]
3 years ago
15

From a stationary position on the ground, I observe a moving boxcar which has a mirror along the front wall but it is open at th

e back of the boxcar. I send a flash of light from my flashlight and time the flash of light as it enters from the back of the boxcar, reflects off the front of the boxcar and returns to the back of the boxcar. Jill (a passenger in the boxcar) also times the round trip of the flash of light. Compare the times recorded on our watches.
A.) The time recorded on her watch is longer.B.) The times recorded on the two watches are the same.C.) The time recorded on her watch is shorter.D.) The answer depends on the reference system you are in.
Physics
1 answer:
Irina-Kira [14]3 years ago
6 0

Answer:

B.) The times recorded on the two watches are the same.

Explanation:

Light moves at a constant speed called the speed of light. In this experiment, the light travels from the back of the moving boxcar, get reflected at the front. Jill who was in the car also noted the time taken for the entry, reflection and exit of the beam of light. The two observers are in different frames of reference.

The times recorded would be the same because light moves at a high speed to cover the required distance. The distance from the back to the front of the boxcar is constant, though it is in motion. The time is recorded when the light travels these paths. The two are in different frames of reference, but the time would be the same.

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A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers
Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

3 0
3 years ago
A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On the w
slamgirl [31]

Answer:

The answer is 3.

Explanation:

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So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.

I hope this answer helps.

4 0
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Should be correct

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