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nexus9112 [7]
3 years ago
10

Explain why a book placed on a table does not move?

Physics
2 answers:
N76 [4]3 years ago
5 0

Answer:

4th: "The force that the book exerts down on the table is equal to the force the table exerts upward"

Explanation:

Basically, a book on a table is balanced in terms of forces. Gravity is opposed by the table and there are no forces acting to move the book in any direction. If there was a breeze that was not strong enough to move the book, it is because the force of friction is equal to that provided by the breeze.

kifflom [539]3 years ago
4 0

Answer:

The force that the book exerts down on the table is equal to the force the table exerts upward.

Explanation:

I am 100% positive. That results in balanced forces which is why it does not move. If you did want it to move, you would apply unbalanced forces.

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The atlas stone’s is a strong man competition where athletes have to load 5 stones of masses 100kg, 120kg, 140kg, 160kg and 180k
lyudmila [28]

Answer: 6250 joules

Explanation:

The work needed to lift an object of mass M by a height H is equal to:

w = M*g*H

where h = 10m/s^2

then the total work that he did is equal to the sum of the work for every stone:

W = (100kg*g*H) + (120kg*g*H) + (140kg*g*H) + (160kg*g*H) + (180kg*g*H)

 = (100kg + 120kg + 140kg + 160kg + 180kg)*g*H

= (500kg)*g*H

and now we can repalce g by 10m/s^2 and H by 125cm

But you can notice that we have two different units of distance, so knowing that 100cm = 1m

we can write H =  125cm = (125/100) m = 1.25 m

Then we have:

H = 500kg*10m/s^2*1.25m = 6250 J

3 0
3 years ago
An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver
Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, E=2.3\times 10^{-10}\ N/C

Vertical distance covered, s=1\ \mu m=10^{-6}\ m

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v^2-u^2=2as

v^2=2as.............(1)

Electric force is F_e and force of gravity is F_g. As both forces are acting in downward direction. So, total force is:

F=mg+qE

F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}

F=4.57\times 10^{-29}\ N

Acceleration of the electron, a=\dfrac{F}{m}

a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}

a=50.21\ m/s^2

Put the value of a in equation (1) as :

v=\sqrt{2as}

v=\sqrt{2\times 50.21\times 10^{-6}}

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

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