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kirza4 [7]
2 years ago
7

What are the forces that act on the ball pushed on the floor​?Plss answer

Physics
1 answer:
vaieri [72.5K]2 years ago
4 0

Explanation:

The ball, for example, will feel gravity pulling it downward and the ground pushing it upward in the direction it is rolling. (Add this if the ball is rolling on the floor.) Friction is the force that causes the ball to slow down because it acts in the opposite direction that it is moving.

If This Answer Helped You Please Mark Me As Brainliest.

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I can't seem to get the right angular acceleration and also not sure how to do part b. Help will be much appreciated.
Tcecarenko [31]

Answer:It’s 5 I believe

Explanation: it says to round to the nearest thousandths, so it’ll be 5.

8 0
3 years ago
You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo
jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

\Delta E_k=W=W_g-W_p

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J

hence, the change in Er is about 1.52J times the initial rotational energy

5 0
3 years ago
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Which of the following situations would cause the greatest decrease in the motion of molecules in a system?
Alina [70]

A. is the right answer since work is negative and Q which is heat in negative also

7 0
3 years ago
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Can someone explain this
Alborosie

Answer:

gravity

Explanation:

That would make the most sense from what I know

8 0
3 years ago
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Short, difficult activities that push your body are called
Reika [66]

Answer: A

Explanation: Any short-duration exercise that is powered primarily by metabolic pathways that do not use oxygen. Examples

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2 years ago
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