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skad [1K]
3 years ago
10

A small bolt with a mass of 41.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire

ction with a frequency of 1.65 Hz. What is the maximum amplitude with which the piston can oscillate without the bolt losing contact with the piston's surface?
Physics
1 answer:
EleoNora [17]3 years ago
8 0

Answer:

Amplitude will be equal to 0.091 m

Explanation:

Given mass of the slits = 41 gram = 0.041 kg

Frequency f = 1.65 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 1.65=10.362rad/sec

Angular frequency is equal to \omega =\sqrt{\frac{k}{m}}

10.362 =\sqrt{\frac{k}{0.041}}

Squaring both side

107.371 ={\frac{k}{0.041}}

k = 4.40 N/m

For vertical osculation

mg=kA

0.041\times 9.8=4.40\times A

A = 0.091 m

So amplitude will be equal to 0.0391 m

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Answer:

Spring constant of the spring will be equal to 9.255 N /m          

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to T=2\pi \sqrt{\frac{m}{K}}, here m is mass and K is spring constant

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0.2245=\sqrt{\frac{0.683}{K}}

Squaring both side

0.0504=\frac{0.4664}{K}

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Answer:

Re=160ohm

Explanation:

Step#1

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Step#2

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Re= (320×320)÷( 320+320)

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Answer:

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17. 3 miles.

Explanation:

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