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Soloha48 [4]
3 years ago
7

A 5.3 kg block initially at rest is pulled to the right along a frictionless, horizontal surface by a constant horizontal force

of magnitude 21 N. Find the block's speed after it has moved through a horizontal distance of 6.4 m.
Physics
2 answers:
Anni [7]3 years ago
6 0
Whatever the person said
dimulka [17.4K]3 years ago
3 0

Answer:

F= MASS *ACCELERATION

12=6*a

a=2 m/sec^2

v^2-u^2=2*a*s

v^2=2*2*3

v=sqrt(12)

v=3.5 m/sec

work done by force= change in kinetic energy

F*S=0.5*M*U^2-0.5*M*V^2

12*3=0 0.5*6*V^2

V=3.46 M/SEC

Explanation:

#if you need any queshtions answered within secs mins hit me up and I got you!

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Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

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E = ¹/₂kA²

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\omega = \sqrt{\frac{k}{m} } \\\\\omega =  \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s

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f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

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