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lesantik [10]
3 years ago
5

A force =(16.8N/s)t is applied to an object of m=45.0kg. Ignoring friction, how far does the object travel from rest in 5.00s?

Physics
1 answer:
kirza4 [7]3 years ago
6 0

Answer:

Explanation:

a = F/m = 16.8t/45 = 0.373*t

v = ∫a*dt = ∫(0.373*t)dt = 0.1865*t²

x = ∫v*dt = 0.1865∫(t²)dt = 0.062*t³

For t = 5.0 sec,

x = 0.062*(5)^3 = 7.75 m

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The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in
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<h3>Force escaping gas exerts</h3>

The force the escaping gas exerts of the rocket is calculated as follows;

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Learn more about force here: brainly.com/question/12970081

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