The answer is:
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"longshore drift" .
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On a planet/celestial body with gravity, the acceleration of gravity is the same for everything. Therefore, all three options are correct.
Hope this helps!
Answer:
Explanation:
A) <em>Large</em>: As she opens her parachute, she begins to displace a large volume of air. This leads to a Large air resistance
B) <em>increase, weight</em>: As she falls, the air resistance force <u><em>increases</em></u>. Now there is a force acting in opposite directions to her <u><em>weight.</em></u>
C)<em>Weight, Decelerate:</em> The skydiver has only the downward force of her <u><em>weight</em></u> pulling down on her, so she starts to <u><em>decelerate</em></u>
D) <em>Weight, Upward, Resultant</em>:
Her <u><em>weight </em></u>is now equal to the <u><em>upward </em></u>force from the ground. Her <u><em>resultant </em></u>force is then zero
E) <em>Increases, same, constant, resultant, terminal</em>:
As she accelerates faster, the air resistance force <u><em>increases</em></u>. It is now the <u><em>same </em></u>as her weight. She now moves at a <u><em>constant</em></u> speed because the <u><em>resultant </em></u>force acting on her is zero. She is now at her <u><em>terminal </em></u>velocity.
F) <em>Increases, same, constant, terminal</em>:
As she decelerates, the air resistance force on her parachute <u><em>increases </em></u>until it is the <u><em>same</em></u> as her weight. She is now moving with a <u><em>constant </em></u>speed until she hits the ground - a new slower <u><em>terminal</em></u> velocity
Electric Energy = Current x Voltage x time = 207 x 9.4 = 1,945.8 J
