Moving from 0m/s to 25m/s in 8.0s equals an average acceleration of...

2 answers:

5 0

Moving from 0m/s to 25m/s in 8.0s equals an average acceleration of 3.125 m/s^2. You get this through the following steps:

25-0/8-0 = 3.125 m/s^2

since the average acceleration (a) = change in vvelocity over change in time

Answer: 3.125 m/s^2

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arlik [135]2 years ago

4 0

<h2>Answer:</h2>

Average acceleration = 3.124 m/s²

<h3>Explanation:</h3>

Given data:

Initial velocity = 0 m/s

Final velocity = 25 m/s

Time = 8.00 s

Average acceleration = ?

Solution:

Average acceleration = (Final velocity - Initial velocity)/time

By putting values in above formula:

Average acceleration = (25 - 0)/8

= 25/8

= 3.125 m/s²

Hence an object moving from 0m/s to 25m/s of velocity in 8.0s has an average acceleration of 3.125 meter per second square. Acceleration is the rate of change of velocity. And average acceleration is equal to change in final velocity and initial velocity divided by time.

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In 1996, NASA performed an experiment called the Tethered Satellite experiment. In this experiment a 4.32 x 104-m length of wire

elena55 [62]

Answer:

Explanation:

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speed, v = 7.07 x 10^3 m/s

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e = B x v x l

e = 5.81 x 10^-5 x 7.07 x 10^3 x 4.32 x 10^4

e = 17745.1 V