Two sumo wrestlers push against each other during a match. One wrestler is much larger than the other. The larger wrestler's feet remain on the floor while the smaller
wrestler's feet slip and he is accelerated in a push right out of the ring. Compare the
force that the large wrestler exerts on the smaller wrestler to the force the smaller
wrestler exerts on the larger wrestler.

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Ответ:

amadileaks

09.06.2020

Physics

The speed as it exits the shower-head openings is 9.6 m/s.

Explanation:

Given that,

Radius $r_{1}= 1.0\ mm$

Number of circular = 20

Radius of pipe $r_{2} = 0.80\ cm$

Speed = 3.0 m/s

We need to calculate the area of circle

Using formula of area of circle

$A=\pi r^2$

We need to calculate the speed

Using equation of continuity

$nA_{1}v_{1}=A_{2}v_{2}$

$v_{1}=\dfrac{A_{2}v_{2}}{nA_{1}}$

Where, A =area of cross section

v = speed

Put the value in the equation

$v_{1}=\dfrac{\pi r_{2}^2\times3.0}{\pi r_{1}^2\times20}$

$v_{1}=\dfrac{\pi\times(0.80\times10^{-2})^2\times3.0}{\pi\times(1.0\times10^{-3})^2\times20}$

$v_{1}=9.6\ m/s$

Hence, The speed as it exits the shower-head openings is 9.6 m/s.

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