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slavikrds [6]
3 years ago
7

I need help. Can you help me? Thank you

Physics
1 answer:
SashulF [63]3 years ago
7 0
What the ionquestion??
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The answer is reminiscence.


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A 65 Kg skier starts at rest at the top of a 150 m long hill that has an incline of 28 degrees. How fast will she be going at th
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Answer:

34 m/s

Explanation:

Potential energy at top = kinetic energy at bottom + work done by friction

PE = KE + W

mgh = ½ mv² + Fd

mg (d sin θ) = ½ mv² + Fd

Solving for v:

½ mv² = mg (d sin θ) − Fd

mv² = 2mg (d sin θ) − 2Fd

v² = 2g (d sin θ) − 2Fd/m

v = √(2g (d sin θ) − 2Fd/m)

Given g = 9.8 m/s², d = 150 m, θ = 28°, F = 50 N, and m = 65 kg:

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v = 33.9 m/s

Rounded to two significant figures, her velocity at the bottom of the hill is 34 m/s.

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If you make an electromagnet using a battery circuit and a steel nail, how can you increase its magnetic strength?
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