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bonufazy [111]
3 years ago
6

How does Newton's three laws of motion apply to the egg drop project

Physics
2 answers:
3241004551 [841]3 years ago
6 0
Answer:
Newton's third law, which states that every action has an equal and opposite reaction, can be related to an egg drop because the forces pushing the egg up and down must follow the law.

Explanation:
Newton's third law states that for each action, there is an equal and opposite reaction.
An egg by itself will splat against the concrete if dropped from a height. The point of an egg drop is to keep this from happening. In other words, you must make sure that the downward force of gravity has an opposite force, great enough to slow the egg down enough to make sure it doesn't break.
If you tie a parachute to save the egg, air drag pushes the parachute up, but the force of gravity on the egg pushes it down still. However, the air drag works opposite gravity and slows the egg's descent enough to ensure it doesn't break on impact.
BabaBlast [244]3 years ago
4 0
<span> Newtons First Law is applied on my egg experiment because it will not move or change it's acceleration until a force acts upon it. Newton's Second Law is applied because of the acceleration caused by natural forces as the egg is plummeting to the earth. And the amount of acceleration the egg has will be largely affected by the amount of force used to hurl the egg to the ground. The Third Law of Motion affects the egg because the amount of force the egg hits the ground with, the ground pushes back with equal and opposite force.</span>
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Given Information:

Power = P = 100 Watts

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Required Information:

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b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

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Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

I = \frac{P}{V}  \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

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Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega

Therefore, the resistance of the bulb is 484 Ω

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