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2 years ago

9

A swimmer pushing off from the wall of a pool exerts a force of 1 newton on the wall. What is the reaction force of the wall on

the swimmer?

1 answer:

8090 [49]2 years ago

3 0

Answer: 1 Newton

Explanation:

"Every action has an equal and opposite reaction."

Please mark as Brainliest if it is correct.

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the rate at which something occurs or is repeated over a particular period of time or in a given sample

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Salmon often jump waterfalls to reach their

natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

$v_x = u cos \theta$

where u is the initial speed and $\theta=37.7^{\circ}$ . The horizontal distance travelled by the salmon is

$d=v_x t = (ucos \theta)t$

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

$t=\frac{d}{v_x}=\frac{d}{u cos \theta}$ (1)

Along the vertical direction, the equation of motion is

$y=h+u_y t -\frac{1}{2}gt^2$

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

$u_y = u sin \theta$ is the vertical component of the initial velocity of the salmon

$g=9.81 m/s^2$ is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

$y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}$

and we can solve this formula for u, the initial speed of the salmon:

$y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s$

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3 years ago

A 12.8-kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 156N and break

Varvara68 [4.7K]

Answer:

$2.4 m/s^2$

Downward

Explanation:

We are given that

Mass of monkey=12.8 kg

Tension=156 N

We have to find the magnitude of the elevator's minimum acceleration.

$T=m(g+a)$

Where $g=9.8 m/s^2$

Substitute the values

$156=12.8(a+9.8)$

$9.8+a=\frac{156}{12.8}=12.2$

$a=12.2-9.8=2.4 m/s^2$

Hence, the acceleration = $a=2.4 m/s^2$

Direction of the elevator's minimum acceleration is downward because the elevator moves downwards.

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3 years ago