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miss Akunina [59]
3 years ago
8

Quentin is playing baseball and is batting. He swings and hits the ball sending it flying to the outfield. Which is greater- the

force of the ball on the bat or the force of the bat on the ball ?
Physics
1 answer:
Bess [88]3 years ago
5 0

Answer:

The force of the ball on the bat is same as the force of the bat on the ball.  

Explanation:

A bat hits the ball and the ball moves to the out filed.

According to the Newton's third law, for every action there is an equal and opposite reaction.

The action and the reaction forces acts on the two different bodies but the magnitude of the force is same.  

As the ball is hitted by the bat, the bat exerts the force on the ball and the same force is exerted on the bat by the ball according to the Newton's third law.

So, the force of the ball on the bat is same as the force of the bat on the ball but the direction of force is opposite.  

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soldi70 [24.7K]

Answer:

The two forces that affect on the pendulum are the force of gravity,

Explanation:

  • ⇒ There is the force of gravity that acts downward upon the bob. It results from the Earth's mass attracting the mass of the bob. And there is a tension force acting upward and towards the pivot point of the pendulum.

Therefore, I hope this helps!

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4 0
2 years ago
The 60.0 kg skier shown below is skiing down a 35.0 degree incline where the magnitude of the friction force is 38.5N
Sunny_sXe [5.5K]

Answer:

a) 4.98m/s²

b) 481.66N

Explanation:

a) Using the Newtons second law of motion

\sum F_x = ma_x\\F_m - F_f = ma_x\\Wsin \theta - F_f = ma_x\\mgsin \theta - F_f = ma_x\\

m is the mass of the object

g is the acceleration due to gravity

Fm is the moving force acting along the plane

Ff is the frictional force opposing the moving froce

a is the acceleration of the skier

Given

m = 60kg

g = 9.8m/s²

\theta = 35°

Ff = 38.5N

Required

acceleration of the skier a

Substituting into the formula;

60(9.8)sin 35^0 - 38.5 = 60a\\588sin35^0 - 38.5 = 60a\\337.26 - 38.5 = 60a\\298.76 = 60a\\a = 298.76/60\\a = 4.98m/s^2\\

Hence the acceleration of the skier is 4.98m/s²

b) The normal force on the skier is expressed as;

N = Wcosθ

N = mgcosθ

N = 60(9.8)cos 35°

N = 588cos 35°

N = 481.66N

Hence the normal force on the skier is 481.66N

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