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DerKrebs [107]
3 years ago
10

A horizontal force of 120 N is required to push a bookcase across the floor at a constant velocity how much is the friction forc

e that acts on the sliding bookcase
Physics
1 answer:
belka [17]3 years ago
8 0

The Friction Force that acts on a sliding bookcase is -120 N

<u>Explanation:</u>

  • Frictional force is the force developed by the two surfaces where they contact and slide against each other. They mainly affected by amount of force impelled them together and by the angle and position of the object.
  •             Frictional Force F = Coefficient of friction * normal force
  • At a constant velocity, there will be no acceleration and the net force will be zero.

               Net Force        = m * a = 0

               120 - Friction   = 0

               Friction Force  = -120 N

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Answer:

The crest to trough distance = 8 m

Explanation:

Given that,

The amplitude of a particular wave is 4.0 m.

We need to find the crest to trough distance.

We know that,

Amplitude = The distance from the base line to the crest or the the distance from the baseline to the trough.

It means,

Distance from crest to trough = 2(Amplitude)

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6 0
3 years ago
The particles that make up protons and neutrons are called
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Answer:

Quark

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Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b
Over [174]

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Explanation:

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Now,

Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }

For Bismuth j=\frac{9}{2} and A is 209.

Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

3 0
3 years ago
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I believe there should be some sort of table attached. Unfortunately I cannot answer this question. Sorry!

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