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sdas [7]
3 years ago
11

A man of weight Wman is standing on the second floor and is pulling on a rope to lift a box of weight Wbox from the floor below.

The rope, which is of negligible weight, has tension T in it, but the pull of the rope is insufficient to lift the box off the floor. What is the magnitude of the normal force that the second floor exerts on the man in this situation?
Physics
1 answer:
slavikrds [6]3 years ago
3 0

Answer:

Explanation:

See the attached figure . See the forces acting on man pulling up the box .

Man is stationary so net force acting on man is zero .

T + R = Wman

R is the reaction force of the ground of second floor  .

R = Wman - T

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Answer:

0.000225 m

Explanation:

Coefficient of linear expansion = \alpha_L=3\times 10^{-6} /C

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Linear expansion

\frac{\Delta L}{L} = \alpha_L\Delta T\\\Rightarrow \Delta L=\alpha_L\Delta T\times L\\\Rightarrow \Delta L=3\times 10^{-6}\times 0.3\times 250=0.000225\ m

Change in length of the Pyrex glass dish is 0.000225 m.

3 0
3 years ago
Bacteria vary in size, but a diameter of 2.0 µm is not unusual. What are the volume (in cubic centimeters) and surface area (in
telo118 [61]

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5 0
3 years ago
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Ghella [55]

Answer:

Pressure on both feet will be 500N/m^2  

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Pressure is equal to ratio of force and area

So pressure P=\frac{F}{A}

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So the pressure on both feet will be 500N/m^2 when person stands on both feet.

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Answer:

v = 384km/min

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You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

d=\frac{1,280,000,000km}{37,000}=34,594.59km

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}

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5 0
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Answer:

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7 0
2 years ago
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