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3 years ago

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Two people carry a heavy electric motor by placing it on a light board 1.80 m long. One person lifts at one end with a force of

350 N, and the other lifts the opposite end with a force of 620 N.
What is the weight of the motor?

1 answer:

bixtya [17]3 years ago

6 0

Answer:

$W=970\ N$

Explanation:

When two people carry a weight placing it on a very light board then we neglect the weight of the board and the board acts like a beam with two mass placed over it and two end point supports are here provided by men.

Then balancing the forces in the vertical direction, weight of the motor is:

$W=350+620$

$W=970\ N$

These men are able to carry the motor in equilibrium condition because they have placed the motor (considering it a point mass) at a position such that the moment due to the lifting forces are equal and opposite.

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Explanation:

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For the pendulum in this problem, we have

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Therefore the period of this pendulum is

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A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

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Answer:

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<u>Known Data</u>

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<u>final momentum</u>

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<u>Impulse</u>

$I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s$

<u>Average Force</u>

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$F=ma$ , so $a=\frac{F}{m_{b}}$ .

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A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

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For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

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Now, we can calculate the final speed of the crate at the end of 10.0 m:

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$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

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$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

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