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lakkis [162]
4 years ago
12

If you push a book against a wall hard enough, it will not slide down even though gravity is pulling it. Use what you know about

friction and Newton’s laws of motion to explain why the book does not fall.
Physics
1 answer:
Rashid [163]4 years ago
3 0

The frictional force is too much compared to the gravitational force

Explanation:

This is because of the great amount of frictional force between the book and the wall.

Also, the force of gravity on the book is not high enough to cause motion of the book to slide downs.

Frictional force is a force that opposes motion.

  • For the book to slide down, it must overcome the force of friction acting on it opposite to the direction of gravity.
  • According to newton's first law 'a body will remain in its state of rest or uniform motion unless acted upon by an external force'.
  • Assuming the external force is gravitational force, clearly we see that the force is less than the frictional exerted on the book by pinning it to the wall.
  • This frictional force is a static friction that is on a body at rest, it must be overcome by a larger force before motion takes place.

           Fr  >  fg

Fr is the frictional force

fg is the gravitational force

learn more:

Newton laws brainly.com/question/11411375

#learnwithBrainly

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The water is flowing through the horizontal constricted pipe. The pressure at one end is 4500Pa, speed is 3m/s and area of cross
Oksana_A [137]

The speed and pressure at another end will be 9 m/sec and -31500 pascals.

<h3>What is gauge pressure?</h3>

The difference between absolute pressure and atmospheric pressure is known as gauge pressure. Relative pressure is another name for gauge pressure.

Given data in problem is;

For horizontal pipe, Z₁=Z₂

Pressure at end 1, P₁ = 4500Pa

Speed at end 1, V₁ = 3 m/sec

Speed at end 2, V₂ = ? m/sec

Area of cross-section at end 1,A₁ =A

Area of cross-section at end 1,A₂ = A/3

From the continuity equation;

A₁V₁=A₂V₂

A× 3  = (A/3)×V₂

V₂ = 9 m/sec

From Bernoulli's equation;

\rm \frac{P_1}{\rho g} + \frac{v_1^2}{2g} +Z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} +Z_2  \\\\ \frac{P_1-P_2}{\rho g } = \frac{v_2^2-v_1^2}{2g} \\\\ \frac{P_1-P_2}{\rho} =  \frac{v_2^2-v_1^2}{2} \\\\ \frac{4500-P_2}{1000} =  \frac{9^2-3^2}{2} \\\\ 4500 -P_2 = 36000 \\\\ P_2 =  - \ 31,500 \ pa

Hence the speed and pressure at another end will be 9 m/sec and -31500 pascals.

To learn more about the gauge pressure, refer to the link;

brainly.com/question/14012416

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