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sertanlavr [38]

3 years ago

12

A bowling ball has a mass of 7.2 kg and a weight of 70.6 N. It moves down the bowling alley at 1 m/s and strikes a pin with a fo

rce of 15.0 N. What is the force that the pin exerts on the bowling ball?
7.2 N
15.0 N
70.6 N
85.6 N

2 answers:

Over [174]3 years ago

6 0

Answer:

15.0 N

Explanation:

Netwon's third law states that:

"When an object A exerts a force on another object B, object B exerts an equal and opposite force on object A"

In this case, the bowling ball is our object A while the pin is the object B. The bowling ball exerts a force of 15.0 N on the pin, therefore according to Newton's third law, the pin exerts a force of 15.0 N on the bowling ball.

Inga [223]3 years ago

4 0

The answer is 15.0N its explained in newtons third law hope this helps:)

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A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

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Answer:

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$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

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$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

&

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

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