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Vlad [161]
3 years ago
15

A book falls of the table and hits the floor. Is work done? Why?​

Physics
2 answers:
morpeh [17]3 years ago
6 0

<u>Yes, work is done when a book falls of the table.</u>

This is because:

When the book falls, it's potential energy is converted into kinetic energy.   As it reaches the floor down, this kinetic energy is converted to heat energy and sound energy due to the impact.

When a force is imposed on an object to cause displacement of that object, work is done on that object. For a force to do work on an object, there should be a displacement and this force should cause the displacement. So here, since the book falls from the table and causes the displacement of the book from the table to the floor. It is said that work is done.

Work can be given by the formula:

W = F • d

where F is the force and d is the displacement.

prohojiy [21]3 years ago
6 0

Answer:

potential energy

Explanation:

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When object is placed between the focus (F) and pole (P) of a concave mirror, magnified and erect image of the object is formed on the back of the mirror.

When object is placed between the centre of curvature and the principal focus of a concave mirror, magnified and inverted image is formed in front of the mirror.

Explanation:

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A child wearing socks is trying to see how far he can slide across his kitchen floor. If
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Answer:

The coefficient of static friction between the ground and the soles of a runner’s shoes is 0.98. What is the maximum speed in which the runner can accelerate without slipping if they have a mass of 73 kg?

Explanation:

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2 years ago
James accelerates his skate board uniformly along a straight road from rest to 10 m/s in 4 seconds. What is James Acceleration?
lawyer [7]

Given:

u(initial velocity)=0

v(final velocity)= 10 m/s

t= 4 sec

Now we know that

v= u + at

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10=0+ax4

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4 0
3 years ago
The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb
Anit [1.1K]

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 in^{2}

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = \frac{F}{A}

⇒ P = \frac{6.4}{0.4}

⇒ P = 16 \frac{lb}{in^{2} }

This is the pressure inside the cylinder.

Let force applied on the brake pad = F_{1}

Area of the brake pad (A_{1})= 1.7 in^{2}

Thus the pressure on the brake pad (P_{1}) =  \frac{F_{1} }{A_{1} }

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = P_{1}

⇒ 16 = \frac{F_{1} }{A_{1} }

⇒ F_{1} = 16 × A_{1}

Put the value of A_{1} we get

⇒ F_{1} = 16 × 1.7

⇒ F_{1} = 27.2 lb

This the total force applied during braking.

The force applied on one wheel = \frac{F_{1} }{4} = \frac{27.2}{4} = 6.8 lb

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7 0
3 years ago
Four forces are exerted on a disk of radius R that is free to spin about its center, as shown above. The magnitudes are proporti
Dmitry_Shevchenko [17]

The given magnitude of forces of F₁ = F₄, F₂ = F₃, F₁ = 2·F₂, give the

forces that exert zero net torque on the disk as the options;

(B) F₂

(D) F₄

<h3>How can the net torque on the disk be calculated?</h3>

The given parameters are;

F₁ = F₄

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F₁ = 2·F₂

Therefore;

F₄ = 2·F₂

In vector form, we have;

\vec{F_4} = \mathbf{\frac{\sqrt{3} }{2} \cdot F_4 \cdot \hat i -  0.5 \cdot F_4 \hat j}

\vec{F_2} = \mathbf{ -F_2 \,  \hat j}

Clockwise moment due to F₄, M₁ = -0.5 \times F_4 \,  \hat j  \times \dfrac{R}{2}

Therefore;

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Counterclockwise moment due to F₂ = -F_2 \,  \hat j  \times \dfrac{R}{2}

Given that the clockwise moment due to F₄ = The counterclockwise moment due to F₂, we have;

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Learn more about the resolution of vectors here:

brainly.com/question/1858958

4 0
2 years ago
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