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MariettaO [177]
2 years ago
9

You are testing a new amusement park roller coaster with an empty car with a mass of 130 kg. One part of the track is a vertical

loop with a radius of 12.0 m. At the bottom of the loop (point A) the car has a speed of 25.0 m/s and at the top of the loop (point B) it has speed of 8.00 m/s.
Physics
1 answer:
vlada-n [284]2 years ago
6 0

Answer:

Work done by friction along the motion is given as

W_f = -5857.8 J

Explanation:

As per work energy theorem we can say

Work done by all forces = change in kinetic energy of the system

so here car is moving from bottom to top

so here the change in kinetic energy is total work done on the car

so here we will have

W_f + W_g = \frac{1}{2}m(v_f^2 - v_i^2)

W_f - mgH = \frac{1}{2}m(v_f^2 - v_i^2)

now plug in all data in it

W_f - (130)(9.81)(2\times 12) = \frac{1}{2}(130)(8^2 - 25^2)

W_f = 30607.2 - 36465

W_f = -5857.8 J

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2 years ago
Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
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A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18
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Answer:

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1)

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v = 18\hat i + 24\hat j + 72\hat k

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natta225 [31]

Answer:

The correct option is;

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