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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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Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

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Explanation:

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