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mixas84 [53]
4 years ago
14

A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor

que of the 6 N force, what is its moment arm?
Physics
1 answer:
Levart [38]4 years ago
6 0

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force F_{1}=6\ N

Second force F_{2}=10\ N

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

\tau=Force\times lever\ arm

So, Here,

\tau_{2}=5 \tau_{1}

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

F_{2}\times d_{2}=5\times F_{1}\times r_{1}

r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}

Where, F_{1}=First force

F_{1}=First force

F_{2}=Second force

r_{1}= distance

Put the value into the formula

r_{2}=\dfrac{5\times6\times0.2}{10}

r_{2}=0.6\ m

Hence, The moment arm is 0.6 m

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Answer:

<h2>The pin's final velocity is 5m/s</h2>

Explanation:

Step one:

given data

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Step two:

The expression for elastic collision is given as

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substituting we have

5*10+2*0=5*8+2*v2

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divide both sides by 2

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v2=5m/s

The pin's final velocity is 5m/s

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