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MaRussiya [10]
3 years ago
12

A 22.8 kg rocking chair begins to slide across the carpet when the push reaches 57.0 N. What is the coefficient of static fricti

on?​

Physics
1 answer:
vfiekz [6]3 years ago
7 0

Answer:

0.255

Explanation:

The following data were obtained from the question:

Force (F) = 57 N

Mass (m) = 22.8 Kg

Coefficient of static friction (µ) =...?

Next, we shall determine the normal reaction (R). This is illustrated below:

Mass (m) = 22.8 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (R) =?

R = mg

R = 22.8 x 9.8

R = 223.44 N

Finally, we can obtain the coefficient of static friction (µ) as follow:

Force (F) = 57 N

Normal reaction (R) = 223.44 N

Coefficient of static friction (µ) =...?

F = µR

57 = µ x 223.44

Divide both side by 223.44

µ = 57/223.44

µ = 0.255

Therefore, the coefficient of static friction (µ) is 0.255.

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3 years ago
Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g
ollegr [7]

Answer:

The  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

<u>velocity of wave on the string with greater tension;</u>

v_1 = \sqrt{\frac{T_1}{\mu }

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} =  \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

v_2^2 = \frac{T_2v_1^2}{T_1}  \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s

<u>Fundamental frequency of wave on the string with greater tension;</u>

<u />f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1}  =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz<u />

Beat frequency = F₂ - F₁

                          = 270.6 - 258

                          = 12.6 Hz

Therefore, the  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

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3 years ago
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Answer:

Strong Nuclear

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