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Shalnov [3]
1 year 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]1 year 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:

The workdone is  W_d =-4400J

Explanation:

The free body diagram is shown on the first uploaded image

From the question we are given that

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Generally  the workdone by the tension force on Magnus is

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This negative sign show that is tension force  is in the opposite direction to Magnus movement (i.e the movement of the bus )

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7 0
2 years ago
. 30
schepotkina [342]

Answer:

Explanation:

Length if the bar is 1m=100cm

The tip of the bar serves as fulcrum

A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum

The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.

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Then, F=1000N

The force acting in the crate lid is 1000N

Option D is correct

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