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Levart [38]
3 years ago
7

A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50

m tall, you observe that it takes the ball 0.210 s to traverse the length of the window. 1)Determine how high above the top of your window the ball was dropped. Ignore the effects of air resistance. (Express your answer to three significant figures.)
Physics
1 answer:
Lelechka [254]3 years ago
5 0

Answer:1.902 m

Explanation:

Given

height of apartment=1.5 m

It takes 0.21 sec to reach the bottom from apartment

So

s=u_1t+\frac{gt^2}{2}

1.5=u_1\times 0.21+\frac{9.81\times 0.21^2}{2}

u_1=6.11 m/s

i.e. if ball is dropped from top its velocity at window is 6.11 m/s

So height of upper floor above window

v^2-u^2=2as

where s= height of upper floor above window

here u=0

6.11^2=2\times 9.81\times s

s=\frac{6.11^2}{2\times 9.81}

s=1.902 m

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Two astronauts, each having a mass of 74.3 kg are connected by a 13.1 m rope of negligible mass. They are isolated in space, orb
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Answer:

  L = 5076.5 kg m² / s

Explanation:

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the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1

as we have two bodies

       L = 2 r m v

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4 0
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In the first part of the problem, light moves from glass to air (n_a=1.00) and the critical angle is \theta_c = 30.8^{\circ}. This means that we can find the refractive index of glass by re-arranging the previous formula:
n_g=n_1 =  \frac{n_2}{\sin \theta_c}= \frac{1.00}{\sin 30.8^{\circ}}=1.95

Now the glass is put into water, whose refractive index is n_w = 1.33. If light moves from glass to water, the new critical angle will be
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