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3 years ago

Newton’s Third Law of Motion states that when one object exerts a force on a second object, the forces are __________________.

1 answer:

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Out of the following choices, Newton's Third Law of Motion states that when one objects exerts a force on a second object, the forces are equal in magnitude and opposite in direction.

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A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th

solniwko [45]

Answer:

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

Explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

1) tangential acceleration which is given as

$a_t = R\alpha$

2) Centripetal acceleration

$a_c = \omega^2 R$

here we know that

$\omega = \alpha t$

$a_c = (\alpha t)^2 R$

now we know that net linear acceleration is given as

$a = \sqrt{a_c^2 + a_t^2}$

so we have

$a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}$

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

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3 years ago

Magnetism is due to the movement of ______.

Ierofanga [76]

Answer:

electrons

Explanation:

Brainliest

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2 years ago

Read 2 more answers

When placed at a certain point, a

Wewaii [24]

Answer: 180

Explanation: Acellus

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2 years ago

A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou

True [87]

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

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3 years ago

How do I do this physics problem about potential energy and kinetic energy?

larisa86 [58]

Ok i apologise for the messy working but I'll try and explain my attempt at logic

Also note i ignore any air resistance for this.

First i wrote the two equations I'd most likely need for this situation, the kinetic energy equation and the potential energy equation.

Because the energy right at the top of the swing motion is equal to the energy right in the "bottom" of the swing's motion (due to conservation of energy), i made the kinetic energy equal to the potential energy as indicated by Ek = Ep.

I also noted the "initial" and "final" height of the swing with hi and hf respectively.

So initially looking at this i thought, what the heck, there's no mass. Then i figured that using the conservation of energy law i could take the mass value from the Ek equation and use it in the Ep equation. So what i did was take the Ek equation and rearranged it for m as you can hopefully see. Then i substituted the rearranged Ek equation into the Ep equation.

So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).

Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.

The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).

I do hope I'm right and i suppose it's better than a blank piece of paper good luck my dude xx

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3 years ago