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Vitek1552 [10]
3 years ago
14

A box is pushed across a horizontal table at constant speed. Of the forces on it, which pair do we know are equal in magnitude b

ecause of Newton's third law? The force of kinetic friction on the box and the force applied to the box by the one pushing it. None of these are interaction pairs. Normal force on the box and the normal force on the table by the box. Normal force on the box and the weight of the box.
Physics
1 answer:
PtichkaEL [24]3 years ago
5 0

Answer:

Normal force on the box and the normal force on the table by the box.

Explanation:

a)The force of kinetic friction on the box and the force applied to the box by the one pushing it:  This pair of forces is compared by Newton's second law.

b)None of these are interaction pairs: If there is a couple of interaction

c)Normal force on the box and the normal force on the table by the box: We know that this pair of forces are equal in magnitude due to Newton's third law, or, the principle of action and reaction, which establishes that when two bodies interact, equal forces and opposite senses appear in each of them.

When a body A exerts a force on another body B, B will react by exerting another force on A of the same modulus and direction although in the opposite direction. The first of the forces is called the action force and the second reaction force :

FAB = - FBA

FAB: It is the force of action of A on B .

FBA: It is the reaction force of B on A

d)Normal force on the box and the weight of the box: This pair of forces is compared by Newton's second law.

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2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her he
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Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, v_s = d/t₂

∴ v_s = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, v_s ≈ 5.79 m/s.

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