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vova2212 [387]
4 years ago
7

If an object starting from rest falls 5.7 seconds how far does it fall

Physics
1 answer:
White raven [17]4 years ago
5 0
Kinematic question!
Start with your knowns!

t = 5.7s \\ v0 = 0 \frac{m}{s}  \\ a =  - 9.8 \frac{m}{ {s}^{2} }
we want to find x!

the only kineamtic that works with this info is the third!
x = v0 \times t + \frac{1}{2}  \times a \times  {t}^{2}
so we plug in our values and get
x = 0 +  \frac{1}{2}  \times ( - 9.8 \frac{m}{ {s}^{2} } ) \times ( {(5.7s)}^{2} )
this'll equal out to -159.2 meters.

our units equal out, so we know our answer is correct!
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4 0
3 years ago
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The question tells you that the distance was increased 3 times the original distance, which means it was moved 3 times

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The motion of a free falling body is an example of __________ motion​
swat32

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accelerated

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A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea
viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

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t' = 5 years

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Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}                           (1)

time taken = 5 years

We know that :

time = \frac{distance}{speed}

5 = \frac{L''}{v}                                                      (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}

Solving the above eqn:

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4 0
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Answer:

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