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OleMash [197]
2 years ago
10

The roller coaster from problem #1 then tops a second hill at 15.0 m/s, how high is the second hill?

Physics
1 answer:
navik [9.2K]2 years ago
4 0

Answer:

68.5

Explanation:

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2 years ago
With what velocity Must a 0.53 kg softball be moving to be equal to the momentum of a 0.31 kg baseball moving at 21 m/s
vladimir2022 [97]

Answer:12.28m/s

Explanation:

momentum of baseball =mass of baseball x velocity of baseball

Momentum of baseball =0.31x21

Momentum of baseball =6.51kgm/s

For a softball to have same momentum with the baseball we can say :momentum of baseball =mass of softball x velocity of softball

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Velocity of softball =6.51/0.53

Velocity of softball =12.28m/s

3 0
3 years ago
A chevrolet corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. what
pychu [463]
By definition we have that the final speed is:
 Vf² = Vo² + 2 * a * d
 Where,
 Vo: Final speed
 a: acceleration
 d: distance.
 We cleared this expression the acceleration:
 a = (Vf²-Vo²) / (2 * d)
 Substituting the values:
 a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
 a = -77268 mi / h ^ 2
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
 First you must make a free body diagram and see the acceleration of the car:
 g = 32.2 feet / sec ^ 2
 a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
 a = -31.48 feet / sec ^ 2
 A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
 A = -22.07 feet / sec ^ 2
 Clearing the braking distance:
 Vf² = Vo² + 2 * a * d
 d = (Vf²-Vo²) / (2 * a)
 Substituting the values:
 d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
 d = 175.44 feet
 answer:
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
5 0
2 years ago
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Nina [5.8K]
It is the photosphere
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