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Greeley [361]
3 years ago
13

frictional force acting on a 0.5 kg object and a floor is 3 N. What is the coefficient of friction between the object and the fl

oor?
Physics
1 answer:
ikadub [295]3 years ago
7 0

Answer:

μ  = 0.6

Explanation:

F =  μN

N = mg

F = μmg

3 N =  μ*0.5 kg * 9.8 m/s²

μ = 3/(0.5*9.8) = 0.6

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A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if
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Answer:

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b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

We have a block-spring system, whose angular frequency (\omega) is defined by the following formula:

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Where:

k - Spring constant, measured in newtons per meter.

m - Mass, measured in kilograms.

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By applying (1) in (2), we get the following formula:

T = 2\pi\cdot \sqrt{\frac{m}{k} }

a) If the mass is doubled, then the period is increased by \sqrt{2}. Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

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