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GuDViN [60]
2 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]2 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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a hippopotamus produces a pressure of 250000 pa when it is standing on all four feet if the weight of the hippo is 40000 N what
mamaluj [8]

0.04m²

Explanation:

Given parameters:

Pressure = 250000Pa

Weight = 40000N

Unknown:

Area of each foot = ?

Solution:

Pressure is the force exerted per unit area of a body

  Pressure = \frac{force}{area}

To find the area;

        Area = \frac{force }{pressure}

    Area = \frac{40000}{250000} = 0.16m²

The force exerted by all the four feet is 0.16m²

the area of each feet = \frac{0.16}{4} = 0.04m²

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A large volume of the solar system's space is occupied by what?
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Junk from our atmosphere
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A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

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