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kobusy [5.1K]
3 years ago
5

Determine the displacement of a plane that is uniformly accelerated from 66 m/s to 88 m/s in 12 s

Physics
1 answer:
german3 years ago
7 0
Displacement is the area under the graph for velocity-time graph so if plane accelerated uniformly from 66m/s to 88m/s the area will be the area of trapezium where formula for area under trapezium is 1/2 (sum of parallel lines) x height so the parallel lines in this case are 66m/s and 88m/s and height is time taken which is 12 so displacement is 1/2 (66+88)(12)= 924m

if you want to know the formula then it is S=1/2 (V+U)(t) where S is displacement V is final velocity U is initial velocity and t is time taken
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The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N). Force is needed to stretch spring is 10.2 N.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

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\rm F_S=Kx \\\\ \rm F_S=8.5 \times 1.2 \\\\ \rm F_S=10.2 N

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brainly.com/question/4291098

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1 year ago
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