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Shalnov [3]
2 years ago
13

A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad

ius is 59 cm. When an object is placed on top of the cylinder, the cylinder compresses by an amount of 5.6 10-7 m. What is the weight of the object?
Physics
1 answer:
maksim [4K]2 years ago
4 0

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

\Upsilon = \frac{F/A}{\Delta L/L_0}

Where,

F = Force/Weight

A = Area

\Delta L= Compression

L_0= Original Length

According to the values given we have to

\Upsilon_{steel} = 200*10^9Pa

\Delta L = 5.6*10^{-7}m

L_0 = 3.2m

r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2

Replacing this values at our previous equation we have,

\Upsilon = \frac{F/A}{\Delta L/L_0}

200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}

F = 38272.5N

Therefore the Weight of the object is 3.82kN

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Answer:

Force applied to smaller cross section is

= 82.63 N

Explanation:

As we know

F_2 A_1 = F_1 A_2

where F1, F2 signifies the weight of the two chair in a hydraulic-lift system

And A_1, A_2 signifies the area of the two respective chairs in a hydraulic-lift system

Given -

F2=1400 N

A1 =1220 Square centimeter

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Substituting the given values in above equation, we get -

1400 * 72 = F1 * 1220\\F2 = 82.63

Force applied to smaller cross section is

= 82.63 N

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Answer:

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We now divide the number of revolution in radians by the time in minutes.

\omega =\frac{11.2\pi}{0.13min}\\\omega=86.15\pi rad/min

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