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Mars2501 [29]
3 years ago
7

A force of 350N is applied to a body. If the work done is 40kJ, what is the distance through which the body moved?

Physics
1 answer:
Studentka2010 [4]3 years ago
7 0

The distance covered by the body is 114.3 m

Explanation:

The work done by a force exerted on an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

For the object in this problem, we have

F = 350 N is the force applied

W=40 kJ = 40,000 J is the work done

\theta=0^{\circ} if we assume that the force is applied parallel to the motion of the object

Solving for d, we find the distance covered by the object:

d=\frac{W}{F cos \theta}=\frac{40,000}{(350)(cos 0)}=114.3 m

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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Explanation:

Given;

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195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

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(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

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