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jek_recluse [69]
2 years ago
15

A rollercoaster loop has a radius of 22.7 m. what is the minimum speed the coaster must have at the top of the loop to not fall

off the track? (unit = m/s)
Physics
1 answer:
Softa [21]2 years ago
4 0

Answer:

v = 14.92 m/s

Explanation:

First, make a <u>free body diagram</u> and see the <u>forces in the y-direction</u>.

  • ∑F_y = F_N - F_g

Use Newton's 2nd Law F = ma to replace ∑F_y with m * a_y.

  • m * a_y = F_N - F_g

The acceleration in the y-direction is the centripetal acceleration, a_c = v^2/r.

  • m * v^2/r = F_N - mg

The <u>normal force is 0</u> because this is where the rollercoaster is not falling off the track yet not touching the track.

  • m * v^2/r = - mg

The masses cancel out.

  • v^2/r = -g

<u>Solve for v</u> to find the speed of the rollercoaster at the top of the loop.

  • v^2 = -(-9.81) * r
  • v = √(9.81 * 22.7)
  • v = 14.9227

The minimum speed the coaster must have at the top of the loop to not fall off the track is <u>14.92 m/s</u>.

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The fraction of the water must evaporate to remove precisely enough energy to keep the temperature constant when water at 37°c has a latent heat of vaporization of lv = 580 kcal/kg is 2.58 times 10 to the minus 3.

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A 38.2 kg wagon is towed up a hill inclined at 17.5 ◦ with respect to the horizontal. The tow rope is parallel to the incline an
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Answer:

v = 8.57 m/s

Explanation:

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So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

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