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bija089 [108]
3 years ago
5

A tug-of-war game is played by five c/hildren: three on one team and two on the other. How much force will the two child team ha

ve to exert against the three child team in order to hold the rope steady if the three children opposing them exert forces of 300 N, 150 N, and 50N respectively? Will the force be equally distributed between the two children?
Physics
1 answer:
dem82 [27]3 years ago
4 0

Answer:

no, 3 porces is more tha 2 so the power between the 3 should be more than 2

Explanation:

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Answer:

Explanation:

Given that,

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And this a velocity V

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Applying conservation of momentum

Momentum is given as p=mv

Initial momentum = final momentum

Po = Pf

(M+7M) × 0 = 7M •V − Mv

0 = 7M•V - Mv

Divide both sides by M

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Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block

The workdone by the 7M mass is

Distance moved by 7M mass is 6.8m, Then, d =6.8m

W = fr × d

Where fr = µkN

When N=W =mg, where m=7M

N= 7Mg

fr = −µk × 7mg

Then, W(7m) = −7µk•Mg×d

W(7m) = −7µk•Mg×6.8

W(7m) = −47.6 µk•Mg

Then, same procedure,

Let distance move by the small mass be m

Work done by M mass

W(m) = −µk•Mg×d'

Since it is a wordone by friction, that is why we have a negative sign.

Using conservation of energy

Work done by 7M mass is equal to work done by M mass

W(7m) = W(m)

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Then, M, g and µk cancels out

We are left with

-46.7 = -d

Then, d = 46.7m

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5 0
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Answer:

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By substituting the values-

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Question 1 (1 point)
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Answer:

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