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2 years ago

7

A footballer kicks a ball from rest. The foot is in contact with the ball for 0.30s and the final velocity of the ball is 15ms-1

.What is the average acceleration of the ball?

1 answer:

Ne4ueva [31]2 years ago

7 0

Answer:

<h2>50m/s^2</h2>

Explanation:

Step one:

given data

initial velocity u= 0m/s since the ball is at rest

time of contact t= 0.3s

final velocity v=15m/s

Required

acceleration a

from the first law of motion

v=u+at

substitute our given data

15=0+a*0.3

15=0.3a

divide both sides by 0.3

a=15/0.3

a=50m/s

<u>The average acceleration is 50m/s^2</u>

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Explanation:

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To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

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$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

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$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

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$A_1 v_1 = A_2 v_2$

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$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

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$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

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B)

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

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C)

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$V=Av$

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Therefore, the volume flow rate is

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Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

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