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GuDViN [60]
3 years ago
13

One observer stands on a train moving at a constant speed, and one observer stands at rest on the ground. The person on the trai

n stands at the back of the car and turns on a flashlight; the light travels to the front of the car, bounces off a mirror, and returns to the back of the car (assume the light is traveling in a vacuum). The observer on the ground measures the time it takes for the light to take this path and knowing the length of the car, calculates the speed of light. What value does she obtain?
a.)greater than c

b.) equal to c

c.) less than c

d.) value cannot be calculated without knowing the speed of the train
Physics
1 answer:
Anna11 [10]3 years ago
4 0

Answer:

Option D.

Value cannot be calculated without knowing the speed of the train

Explanation:

The speed of the beam can only be calculated accurately when the speed of the train is put into consideration. Based of the theory of relativity, the observer is on the ground, and the train is moving with the beam of light inside it. This causes a variation in the reference frames when making judgements of the speed of the beam. The speed of the beam will be more accurate if the observer is moving at the same sped of the train, or the train is stationary.

To get the correct answer, we have to subtract the speed of the train from the speed calculated.

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Andrews [41]

The work done by the applied force on the block against the frictional force is 15.75 J.

<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

  • F is applied force
  • Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

F = 9.86 N

W = Fdcosθ

W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

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5 0
2 years ago
What is the area under the curve for the histogram below?
olga2289 [7]
Well i honestly wouldn’t know
5 0
3 years ago
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An 87.6 g lead ball is dropped from rest from a height of 7.00 m. The collision between the ball and the ground is totally inela
bija089 [108]

Taking specific heat of lead as 0.128 J/gK = c

We have energy of ball at 7.00 meter height = mgh = 87.6*10^{-3}*9.81*7

When leads gets heated by a temperature ΔT energy needed = mcΔT

                                                                      = 87.6*10^{-3}*0.128*10^3ΔT

Comparing both the equations

                      87.6*10^{-3}*9.81*7 = 87.6*10^{-3}*0.128*10^3ΔT

                        ΔT = 0.536 K

                        Change in temperature same in degree and kelvin scale

                                      So ΔT = 0.536 ^0C

7 0
3 years ago
Question 15 of 15
castortr0y [4]

Answer:

a and b

Explanation:

are the correct answers

7 0
3 years ago
Need help please!! I literally am dumb and can’t do this
natta225 [31]

Answer:

The answer should be A and B. I don't fully know so don't quote me on that! Haha.

Explanation:

6 0
3 years ago
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