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BlackZzzverrR [31]
3 years ago
5

Truck drivers approaching a steep hill that they must climb often increase their speed. What good does this do, if any?

Physics
1 answer:
Inga [223]3 years ago
6 0
Um, this doesn't make any sense. By climbing a hill, you are decreasing your momentum and kinetic energy, so it slows you down. The only positive, is after you have climbed the hill, you have more potential energy, and it will be released once you go down the hill, but you will not be as fast as if you ignored the hill.
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Answer:

F = \frac{-Gm_{1}m_{2} }{r^{2} }.

Explanation:

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F = \frac{-Gm_{1}m_{2} }{r^{2} }

Here ,m_{1} = 2m

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Thus , F = \frac{-G.2m.\frac{m}{2} }{r^{2} }

          F = \frac{-Gm_{1}m_{2} }{r^{2} }.

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valina [46]
The  spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
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f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz
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In a spring-mass system, the maximum velocity of the object is given by
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v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s
6 0
3 years ago
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Answer:

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Explanation:

we need it to answer

4 0
3 years ago
Read 2 more answers
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