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Gala2k [10]
3 years ago
13

A baseball thrown by a pitcher is hit by a batter. At the moment when the ball hits the bat, the force exerted on the bat by the

ball
is Fball on bat and the force exerted on the ball by the bat is Fbat on ball The relationship between these forces is

A. Fball on bat = -Fbat on ball
B. Fball on bat = -Fbat on ball

C. Fball on bat < Fbat on ball

D. Fball on bat > Fbat on ball
Physics
1 answer:
Slav-nsk [51]3 years ago
8 0

Newton's Laws of motion describe the motion of an object based on the applied force

The correct option that gives the relationships between the forces is the the option;

\underline{F_{ball}}<u> on bat</u> = \underline{F_{bat}}<u> on ball</u>

<em>(Either option A or B without the minus symbol before </em>F_{bat}<em>, likely typographical error)</em>

<em />

Reason:

Force exerted on the bat by the ball = F_{ball}

Force exerted on the ball by the bat = F_{bat}

Given that the batter hits the ball with the bat, the force exerted by the bat on the ball, F_{bat}, is reacted to by the force of the ball acting on the bat, F_{ball}

According to Newton's third law of motion, action and reaction are equal and opposite. Therefore, at the moment when the ball hits the bat, we have;

\underline{F_{ball}}<u> on bat</u> = \underline{F_{bat}}<u> on ball</u>

<em>Where the force of the bat is high, the ball is accelerated to travel at high speed</em>

Learn more Newton's Laws of motion here:

brainly.com/question/24522313

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A child goes down the slide,starting from rest. If the length of the slide is 2m and it takes the child 3 seconds to go down the
lapo4ka [179]

Answer:

0.44 m/s^2

Explanation:

We have the following data:

- distance covered by the child: d = 2 m (length of the slide)

- time taken to cover this distance: t = 3 s

- initial velocity of the child: 0 m/s (he starts from rest)

So we can find the acceleration by using the equation:

d=ut+\frac{1}{2}at^2

Where a is the acceleration.

Substituting the values and solving for a,

a=\frac{2d}{t^2}=\frac{2(2)}{3^2}=0.44 m/s^2

3 0
3 years ago
A basketball is tossed up into the air, falls freely, and bounces from the wooden floor. From the moment after the player releas
Tju [1.3M]

Answer:

Tha ball- earth/floor system.

Explanation:

The force acting on the ball is the force of gravity when ignoring air resistance. At the moment the player releases the ball, until it reaches the top of its bounce, the small system for which the momentum is conserved is the ball- floor system. The balls exerts and equal and opposite force on the floor. <u>Here the ball hits the floor, because in any collision the momentum is conserved. Moment of the ball -floor system is conserved</u>. Mutual gravitation bring the ball and floor together in one system. As the ball moves downwards, the earth moves upwards, although with an acceleration on the order of 1025 times smaller than that of the ball. The two objects meet, rebound and separate.

5 0
3 years ago
A 60 kg sprinter has a momentum of +600 kg-m/s when he crosses the finish
MakcuM [25]

Answer:

10 ms⁻¹

Explanation:

The amount of momentum that an object has is dependent upon two factors

  • mass of the moving object  
  • speed of motion

In terms of an equation,

Momentum (P) = Mass(m)×velocity(v)

                     P = m×v

                 600 = 60 × v ⇒ v = 10 ms⁻¹

3 0
3 years ago
All of the waves in the electromagnetic spectrum are _______ waves.
crimeas [40]
<span>All of the waves in the electromagnetic spectrum are transverse waves.</span>
5 0
3 years ago
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Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh
mel-nik [20]

Answer:u=\frac{v}{2}\sqrt{5-4sin\phi }

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

m(2v)+m(-vsin\phi )=2m(ucos\theta )

2ucos\theta =v(2-sin\theta )------1

Conserving momentum in x direction

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squaring and adding 1 &2

(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2

4u^2=4v^2+v^2-4v^2sin\phi

4u^2=5v^2-4v^2sin\phi

u=\frac{v}{2}\sqrt{5-4sin\phi }

7 0
3 years ago
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