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Radda [10]
3 years ago
7

A 26.0 kg child plays on a swing having support ropes that are 2.40 m long. A friend pulls her back until the ropes are 45.0 ∘ f

rom the vertical and releases her from rest. What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing? How fast will she be moving at the bottom of the swing? How much work does the tension in the ropes do as the child swings from the initial position to the bottom?
Physics
1 answer:
Sloan [31]3 years ago
8 0
A)Ep'=mgh=mgl(1-cosa).At the bottom of the swing Ep=0(reference level),so the potential energy as the child is just released is bigger than the potential energy at the bottom of the swing.;B)The speed of the child at the bottom of the swing-->v=√(2gh)=√[2gl(1-cosa)];C)I don't think that the tension does any work.
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P\alpha T^4

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then we can make a ratio between the state of before quadruple (with subscript 1) and after (with subscript 2) as:

\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{T_{2}^4}

As

T_{2}=4T_{1}

Then

\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{(4T_{1})^4}

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7 0
3 years ago
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worty [1.4K]

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Explanation:

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Where, P_{in} =inside Pressure

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Put the value into the formula

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6 0
3 years ago
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tamaranim1 [39]
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Problem:
pav-90 [236]

Answer:

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