Question

1) A 1650-kg car is traveling at 37.6 m/s. How fast must a 91.0-kg bicycle and rider be traveling to have the same kinetic energy as the car? 2) It takes 209 kJ of work to accelerate a car from 20.1 m/s to 28.4 m/s. What is the car's mass? 3)A pole-vaulter just clears the bar at 4.84 m and falls back to the ground. The change in the vaulter's potential energy during the fall is -3000 J. What is his weight?

Name: 1 a 1650 kg car is traveling at 376 ms how fast must a 910 kg bicycle and rider be traveling to have the same kinetic energy as the car 2 it takes 209 kj of work to accelerate a car from 201 ms to 284
Uploaded: 2023-10-10T00:23:21-08:00
Duration: 3 min 57 s
Channel: Nishant Kumar
Description: 1 a 1650 kg car is traveling at 376 ms how fast must a 910 kg bicycle and rider be traveling to have the same kinetic energy as the car 2 it takes 209 kj of work to accelerate a car from 201 ms to 284

          1) A 1650-kg car is traveling at 37.6 m/s. How fast must a 91.0-kg bicycle and rider be traveling to have the same kinetic energy as the car?

2) It takes 209 kJ of work to accelerate a car from 20.1 m/s to 28.4 m/s. What is the car's mass?


3)A pole-vaulter just clears the bar at 4.84 m and falls back to the ground. The change in the vaulter's potential energy during the fall is
-3000 J. What is his weight?

Added by Renata C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Nishant Kumar

10/10/2023

Step 1

We can set the kinetic energy of the car equal to the kinetic energy of the bicycle and rider and solve for the velocity of the bicycle and rider. 1/2 * 1650 kg * (37.6 m/s)^2 = 1/2 * 91.0 kg * v^2 Solving for v, we get v = sqrt[(1650 kg * (37.6 m/s)^2) / 91.0 Show more…

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1) A 1650-kg car is traveling at 37.6 m/s. How fast must a 91.0-kg bicycle and rider be traveling to have the same kinetic energy as the car? 2) It takes 209 kJ of work to accelerate a car from 20.1 m/s to 28.4 m/s. What is the car's mass? 3)A pole-vaulter just clears the bar at 4.84 m and falls back to the ground. The change in the vaulter's potential energy during the fall is -3000 J. What is his weight?