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MrMuchimi
3 years ago
6

A weightlifter lifts a set of weights a distance of 2.00 m. If the weights have a mass of 60.0 kg and are pushed vertically upwa

rd at a constant speed, how much work is done on the weights?
Physics
2 answers:
Sergio [31]3 years ago
6 0
The amount of work done on the weights would be approximately 15.0 Kg of work
leonid [27]3 years ago
5 0

Answer:

1200Nm

Explanation:

Work is said to be done if a force applied to an object causes the object to move through a distance. Mathematically,

Work done = Force × Distance

Given distance = 2m

Mass of the body = 60kg

The weight W of the bodies =mass × acceleration due to gravity

W = 60×10 = 600N

Work done = 600×2

Work done = 1200Nm

1200Nm of work is done on the weights

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The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

       \text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}

Similarly,

      \text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}

We know the below formulas,

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          \text {volume of wire}=\pi \times r^{2} \times l

When equating both the equations, we can find length of wire as below, where \pi=\frac{22}{7}

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         \frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l

The \pi value gets cancelled as common on both sides, we get

           \frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l

The 10^{-6} value gets cancelled as common on both sides, we get

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3 years ago
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Hope this helps

plz mark as brainliest!!!!!!!

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Answer:

answer

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