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4 years ago

A batter swings at a baseball. The action force is the bat hitting the ball with a force of 5N. What is the reaction force?

Physics

2 answers:

nlexa [21]4 years ago

7 0

Answer:

The ball hitting the bat.

Explanation:

It is given that, a batter swings at a baseball. When the ball is being hitted by the bat, the action force acting on it is 5 N. Let R is the reaction force. It is equal to the action force but it acts in the opposite direction as per Newton's third law of motion.

The action and reaction forces acts on two different objects. Here, if the action force is " bat hitting the ball" then the reaction force will be "the ball hitting the bat" with a force of -5 N.

Hence, the correct option is (d) "the ball hitting the bat".

omeli [17]4 years ago

6 0

The ball hitting the bat

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A 225 kg block is pulled by two horizontal forces. The first force is 178 N at a 41.7-degree angle and the second is 259 N at a

yawa3891 [41]

Answer:

52.9 N, 364.7 N

Explanation:

First of all, we need to resolve both forces along the x- and y- direction. We have:

- Force A (178 N)

$A_x = (178 N)(cos 41.7^{\circ})=132.9 N\\A_y = (178 N)(sin 41.7^{\circ})=118.4 N$

- Force B (259 N)

$B_x = (259 N)(cos 108^{\circ})=-80.0 N\\B_y = (259 N)(sin 108^{\circ})=246.3 N$

So the x- and y- component of the total force acting on the block are:

$R_x = A_x + B_x = 132.9 N - 80.0 N =52.9 N\\R_y = A_y + B_y = 118.4 N +246.3 N = 364.7 N$

7 0

3 years ago

An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. How high does it rise (v = 0 cm

Anit [1.1K]

Answer:

The value of d is 20.4 m.

(C) is correct option.

Explanation:

Given that,

Initial velocity = 20 m/s

Final velocity = 0

We need to calculate the time

Using equation of motion

$v = u+gt$

Where, u = Initial velocity

v = Final velocity

Put the value into the formula

$t = \dfrac{20}{9.8}$

$t= 2.04\ sec$

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

$s=0+\dfrac{1}{2}\times9.8\times(2.04)^2$

$s=20.4\ m$

Hence, The value of d is 20.4 m.

4 0

4 years ago

When jogging outside you accidently bump into a curb. Your feet stop but your body continues to move forward and you end up on t

oee [108]

Inertia I think because I've heard it around school and in science

6 0

3 years ago

What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m

masya89 [10]

Answer: $0.0058 m/s^{2}$

Explanation:

Centripetal acceleration $a_{C}$ is calculated by the following equation:

$a_{C}=\frac{V^{2}}{r}$

Where:

$V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s$ is the Earth's orbital speed

$r=1.5(10)^{11} m$ is the orbital radius

$a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}$

$a_{C}=0.0058 m/s^{2}$

4 0

3 years ago

A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire

oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

6 0

3 years ago