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Galina-37 [17]
3 years ago
11

2. A woman prevents a 3kg brick from falling by pressing it against a vertical wall. The coefficient of friction

Physics
1 answer:
Anettt [7]3 years ago
6 0
<h3>Answer:</h3>

49 N

<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of the brick as 3 kg
  • The coefficient of friction as 0.6

We are required to determine the force that must be applied by the woman so the brick does not fall.

  • We need to importantly note that;
  • For the brick not to fall the, the force due to gravity is equal to the friction force acting on the brick.
  • That is; Friction force = Mg

But; Friction force = μ F

Therefore;

μ F = mg

0.6 F = 3 × 9.8

0.6 F = 29.4

      F = 49 N

Therefore, she must use a force of 49 N

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Which metals have similar chemical properties? Check all that apply.
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A student grabs a piece of paper and tears it from the middle. Which statement best explains the forces involved?
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b. The normal force between the molecules of the paper is overcome by the contact force of the hands.

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A thin 1.5 mm coating of glycerine has been placed between two microscope slides of width 0.8 cm and length 3.9 cm . Find the fo
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The  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

<h3>Force required to pull one end at a constant speed</h3>

The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;

F = ma

where;

  • m is mass
  • a is acceleration

At a constant speed, the acceleration of the object will be zero.

F = m x 0

F = 0

Thus, the  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

Learn more about constant speed here: brainly.com/question/2681210

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A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bum
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Hits per second=199 hit/s

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#Given the angular velocity, \omega=33\frac{1}{3} rev/min , radius of the record r=0.1m and the distance between any two successive bumps on the groove as d=1.75mm.

The linear speed of the record in meters per second is:

v=\omega r=33\frac{1}{3}\times\frac{2\pi}{60}\times 10\times 10^{_2}\\\\=0.3843m/s\\

#From v above, if the bumps are uniformly separated by 1m, then the rate at which they hit the stylus is:

Hits/second=v/d    \ \ \ \ d=1.75mm\\\\=0.3483/0.000175\\\\=199.0385714\approx 199

Hence the bumps hit the stylus at around 199hit/s

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