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Ratling [72]
3 years ago
7

A 30.0 kg object rests on a flat, frictionless surface. A rope lifts up on the object with a force of 309 N. What is the acceler

ation of the object?
Physics
1 answer:
saveliy_v [14]3 years ago
4 0
Total force = Frope - mg = 309-300
total force = ma
9 = 30a
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Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

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Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

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mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

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Second Trial

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Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

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Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

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