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Shkiper50 [21]
3 years ago
12

A 12000 kg railroad car is traveling at a 2 m/s when it strikes another 10000 kg railroad is at rest. If the car lock together w

hat is the final speed of the two
Physics
1 answer:
zimovet [89]3 years ago
5 0

The final speed is 1.09m\frac{m}{s}

Why?

To calculate the final speed, we need to know that we are working with an inelastic collision, it means that both bodies stick together after the colission. We can calculate the final speed of the bodies (if exists) using the following equation:

m_{1}v{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{f}

We are given the following information:

m_{1}=12000kg\\v_{1}=2\frac{m}{s}\\m_{2}=10000kg\\v_{2}=0\frac{m}{s}(at rest)

So, substituting and calculating, we have:

m_{1}v{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{f}\\\\12000kg*2\frac{m}{s}+10000kg*0\frac{m}{s}=(12000kg+10000kg)*v_{f}\\\\v_{f}=\frac{24000\frac{kg.m}{s} }{22000kg}=1.09m\frac{m}{s}

Hence, the final speed is 1.09 m/s.

Have a nice day!

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Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more t
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The force result in stretching the spring 10.0 centimeters is 2.5N.

<h3>What is Hooke's law?</h3>

If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.

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It might be to late but the answer is C
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