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Amanda [17]
3 years ago
12

A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min.

What is the magnitude of the force needed to keep the car moving constant speed if friction is negligible?
Physics
1 answer:
dedylja [7]3 years ago
5 0

Answer:

F = 768 N                  

Explanation:

It is given that,

Speed of the elevator, v = 3.2 m/s

Grain drops into the car at the rate of 240 kg/min, \dfrac{dm}{dt}=240\ kg/min = 4\ kg/s

We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

F=\dfrac{dp}{dt}

F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}

Since, the speed is constant,

F=m\dfrac{dv}{dt}

F=v\dfrac{dm}{dt}

F=3.2\times 240

F = 768 N

So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.

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<h3>Answer:</h3>

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