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LenKa [72]
3 years ago
11

1. A car’s velocity increases from 4.0 m/s to 36 m/s over a 4.0 second time interval. What is its average acceleration?

Physics
1 answer:
Finger [1]3 years ago
7 0

Explanation:

36-4/4= 9 m/squared. meter per squared.

acceleration unit is meter per second Square.equation is velocity by time.for average final(36) minus initial(4)

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Bob is watching Anna fly by in her new high-speed plane, which Anna knows to be 60 m in length. As a greeting, Anna turns on two
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Answer:

0.196*c

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  dt = y (dt' + \frac{v*dx}{c^2} ) \\\\dt' = 0\\dx' = 60 m\\\\t_2 - t_1  = \sqrt{\frac{1}{(1 - v^2/c^2)  } } * (\frac{v*(x_2 - x_1)}{c^2} )

we are assuming that dx' > 0 , hence we are saying that light emitted from front end of the plane is event number two and that coming from the back end is event number one. As the time in Bob's frame turns out to be positive, this means that event number two happened after event number one in his frame, in other words, the light emitted from the back end of Anna's  plane arrives first to Bob's eyes.

Given:

dt' = 40ns

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dt'  = \sqrt{\frac{1}{(1 - v^2/c^2)  } } * (\frac{v*dx'}{c^2} )\\\\\sqrt{1 - \frac {v^2}{c^2  } } = (\frac{v*dx'}{c^2*dt} )\\\\v^2 = \frac{c^4*dt^2}{c^2*dt^2 + dx'^2}\\\\v = \frac{c^2*dt}{\sqrt{c^2*dt^2 + dx'^2} }\\\\v = \frac{(3*10^8)^2*40*10^9}{\sqrt{(3*10^8)^2*(40*10^9)^2 + 60^2} }\\\\v = \frac{3.6*10^9}{61} \\\\v = 0.196 c

8 0
3 years ago
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