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3 years ago

14

At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri

ze. The ball bounces straight back at 30% of its incoming speed, knocking the bottle straight forward.What is the bottle's speed, as a percentage of the ball's incoming speed?Express your answer using two significant figures.

1 answer:

vfiekz [6]3 years ago

6 0

Answer:

$6.5\%\ of\ u_1$

Explanation:

$m_{1}$ = Mass of ball = 0.12 kg

tex]m_{2}[/tex] = Mass of bottle = 2.4 kg

$u_{1}$ = Initial velocity of ball

$u_{2}$ = Initial velocity of bottle

$v_{1}$ = Final velocity of ball = $-0.3u_{1}$

$v_{2}$ = Final velocity of bottle

As linear momentum is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow 0.12u_{1}+2.4\times 0=-0.12(0.3u_{1})+2.4v_{2}\\\Rightarrow 0.156u_1=2.4v_2\\\Rightarrow \frac{v_2}{u_1}=0.065\\\Rightarrow v_2=0.065\times 100\times u_1\\\Rightarrow v_2=6.5\%\ of\ u_1$

The bottle's speed is $6.5\%\ of\ u_1$

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Given the following data;

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Substituting into the equation, we have;

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Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

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Answer:

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Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

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$E_1=K\dfrac{q_1}{r_1^2}$

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Now by putting the va;ues

a)

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

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$E_2=21.58\times 10^{6}\ N/C$

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As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

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$E_2=147.92\times 10^{6}\ N/C$

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