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3 years ago

7

Where would you find the least dense water? at 1,000 m in the Antarctic Ocean ,at 200 m in the Arctic Ocean, at the surface of t

he Atlantic Ocean, below 1,000 m in the Pacific Ocean

1 answer:

VladimirAG [237]3 years ago

6 0

Answer:

At the surface of the Atlantic Ocean

Explanation:

The atlantic ocean is a worm ocean and the surface has the least dense water. This is because the denser layer of water sinks, while the least dense floats.

The Atlantic Ocean stands out in the choices with warmer temperatures at depths. This results from the different flow dynamics in the semi-enclosed Mediterranean Sea.

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Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

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<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

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Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :

$V_{2}=V_{1}+at$ (6)

$V_{2}^{2}=V_{1}^{2}+2ad$ (7)

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$a$ is the acceleration

$d=0.01 m$ is the length the ball was compressed

$t$ is the time

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$a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}$ (8)

$a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}$ (9)

$a=90.25 m/s^{2}$ (10)

Substituting (10) in (6):

$57 m/s=-38 m/s+(90.25 m/s^{2})t$ (11)

Finding $t$ :

$t=1.052 s$ (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force $F$ is proportional to the variation of momentum $\Delta p$ in time $\Delta t$ :

$F=\frac{\Delta p}{\Delta t}$ (13)

$F=\frac{65.075 kgm/s}{1.052 s}$ (14)

Finally:

$F=61.82 N$

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