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zalisa [80]
2 years ago
13

A skateboarder drops in off the top of one side of the half pipe shown below. She does not push off and starts from rest. She st

ands straight as she skates down one side and up the other. She expected to get to the top of other side but didn’t make it. The skateboarder recalled the law of conservation of energy from science and didn’t understand why she didn’t make it to the top of the other side.
Which of the following actions would help the skater reach the top of the other side of the half pipe?


Lubricate the wheels in order to reduce frictional force that is causing the skateboard to lose velocity and bend down while skating off the side and bottom of the half pipe.


Hold a weight in order to increase her mass and cause the skateboard to move with a greater velocity down the half pipe.


Sit on the skateboard in order to lower her center of mass and increase the potential energy of the skateboard.


Push off at the bottom of the half pipe in order to replace lost energy and add kinetic energy to the skateboard.
Physics
1 answer:
solong [7]2 years ago
4 0

Answer:

v

Explanation:

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borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

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period of oscillation, T = 0.13 s

First, determine the spring constant, k;

T = 2\pi \sqrt{\frac{m}{k} }

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T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

Now, determine the amplitude of oscillation, A;

E = \frac{1}{2} kA^2

where;

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A is the amplitude of the oscillation

E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

Therefore, the amplitude of the oscillation is 2.82 cm

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