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Goryan [66]
2 years ago
9

At night, large bodies of water release heat to the atmosphere quickly. True False

Physics
1 answer:
Nikolay [14]2 years ago
3 0

Answer:

false I hope. have a good day!

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Use the data provided to calculate the gravitational potential energy of each cylinder mass. Round your answers to the nearest t
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Explanation: just did it on Edge

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Longitude lines are not ____________________________ like latitude lines.
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They are not parallel like latitude lines.  They meet at the poles, but latitude lines never meet
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Convert:<br> 1) 2kg into gram<br> 2) 5200m into km<br> 3) 20cm into m
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Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g
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Answer: a) It will take more time to return to the point from which it was released

Explanation: To determine how long it takes for the ball to return to the point of release and considering it is a free fall system, we can use the given formula:

d=v_{0}.t + \frac{1}{2} .a.t^{2}, where:

d is the distance the ball go through;

v₀ is the initial velocity, which is this case is 0 because he releases the ball;

a is acceleration due to gravity;

t is the time necessary for the fall;

Suppose <em>h</em> is the height from where the ball was dropped.

On Earth:

h=0.t + \frac{1}{2}.10.t^{2}

h = 5t²

t_{T} = \sqrt{\frac{h}{5} }

On the other planet:

h =  0.t + \frac{1}{2}.30.t^{2}

h = 15.t²

t_{P} = \sqrt{\frac{h}{15} }

Comparing the 2 planets:

\frac{t_{T} }{t_{P} } = \frac{\sqrt{\frac{h}{5} } }\sqrt{{\frac{h}{15} } }

\frac{t_{T} }{t_{P} } = \sqrt{3}  or t_{T} = \sqrt{3}.t_{P}

Comparing the two planets, on the massive planet, it will take more time to fall the height than on Earth. In consequence, it will take more time to return to the initial point, when it was released.

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