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Murljashka [212]
3 years ago
12

Your cat Goldie (mass 8.50 kg ) is trying to make it to the top of a frictionless ramp 39.930 m long and inclined 27.0 ∘ above t

he horizontal. Since Goldie can’t get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant 37.40 N force parallel to the ramp (there is a set of stairs alongside the frictionless ramp on which you can walk). If Goldie is moving at 2.000 m/s at the bottom of the ramp (assume she has gotten a running start), what is her speed when she reaches the top of the incline?
Physics
1 answer:
Citrus2011 [14]3 years ago
7 0

Answer:

The speed when se reaches the top of the incline is 0.28 m/s

Explanation:

The work done is equal to the change of kinetic energy, then:

Wg + Wf + Wn = ΔEk

Where

Wg = work done by gravity

Wf = work done by force

Wn = work done by normal force

-mgdsin\theta +Fd+0=\frac{1}{2} *m*(v_{2} ^{2} -v_{f} )

Where

m = 8.5 kg

g = 9.8 m/s²

d = 39.93 m

F = 37.4 N

vf = 2 m/s

Replacing:

39.93*(-8.5*9.8*sin27+34.4)=\frac{1}{2} *8.5*(v_{2} ^{2} -2^{2} )\\v_{2} =0.28m/s

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3 years ago
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3 years ago
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8 0
4 years ago
answer questions1) The south pole of the steel magnet will be brought closer to the iron nail head. Which pole will be at the po
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Answer:

North pole

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8 0
1 year ago
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Tcecarenko [31]

Answer:

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Explanation:

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Here the capacitance is 11 F for half and fully charged capacitor.

4 0
3 years ago
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