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cluponka [151]
2 years ago
5

2 part question

Physics
1 answer:
schepotkina [342]2 years ago
8 0
the answer is yes number5 hope this helps did before
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Viefleur [7K]
(3) The frictional force exerted by the floor on the box
4 0
3 years ago
Can anyone plz answer this now and give me a right answer?
klemol [59]

Answer:

90 degree hope it help :))

5 0
2 years ago
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If the net torque is zero, what does this imply about the clockwise and counterclockwise torques.
lapo4ka [179]

Answer:

Explanation:

When two forces acting on a line of action and they are equal in magnitude but opposite it direction, it forms a couple.

Torque is defined as the product of either force and the perpendicular distance between the two forces.

It is a vector quantity.

The net torque is zero, it means the anticlockwise torque is equal to the clockwise torque.

It means they balances each other.

6 0
3 years ago
Which of the following pairings are more likely to be held together with the strong nuclear force?
aivan3 [116]

Which of the following pairings are more likely to be held together with the strong nuclear force

Explanation:

1.What does a strong nuclear force do in an atom? It repels electrons from other electrons. It repels protons from other protons. It attracts protons and neutrons.

2.The chain reaction requires both the release of neutrons from fissile isotopes undergoing nuclear fission and the subsequent absorption of some of these neutrons in fissile isotopes.

3.The strong nuclear force holds most ordinary matter together because it confines quarks into hadron particles such as the proton and neutron. In addition, the strong force binds these neutrons and protons to create atomic nuclei.

8 0
3 years ago
Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the gr
DiKsa [7]

Answer:

(2/3) times height collision occur

Explanation:

for ball A

from the kinematic equation

the distance of ball A is

x_A = v_0 t + \frac{1}{2} at^2

v_0* t= velocity ( time ) = distance

since ball is at height, the above equation changes as

x_A = H - \frac{1}{2} gt^2

for ball B

xB = v0 t - \frac{1}{2} gt^2

the condition of collision is

xA = xB

vA = - 2vB (given)

from the kinematic equation

the speed of the ball A is

v_A = u- gt

since initial speed of the ball A is zero

, so

v_A = -gt

the speed of the ball B is

v_B = v_0 - gt

sincev_A = - 2v_B

   -gt = -2 ( v_0 - gt)

-gt = -2 v_0 +2gt

3gt =2 v_0

t = \frac{2v_0}{3g}

since x_A = x_B

H -  \frac{1}{2} gt^2= v_0 t - \frac{1}{2} gt^2

H = v_0 t

= v_0 (2v_0/3g)

= \frac{2 v_0^2}{ 3g}

x_A = 2 (\frac{v_0^2}{ 3g})- \frac{1}{2} gt^2

  = 4 \frac{v_0^2}{9g} = (2/3) H

so, (2/3) times height collision occur

3 0
3 years ago
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