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3 years ago

EXPERTS/ACE/GENIUS (and people that wanna actually help)

2 answers:

8 0

Hello! Force = mass * acceleration. We can go ahead and eliminate A and B, because those answers are too small. We can multiply 4.5 (mass) * 9 (acceleration) to find the amount of force needed. 4.5 * 9 is 40.5.The amount of force needed is 40.5 N. The answer is C: 40.5 N.

erma4kov [3.2K]3 years ago

3 0

I just took the test and the answer was C. 40.5n. hope this helps everyone who was looking for this answer

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VashaNatasha [74]

What a delightful little problem !
Here's how I see it:

When 'C' is touched to 'A', charge flows to 'C' until the two of them are equally charged. So now, 'A' has half of its original charge, and 'C' has the other half.

Then, when 'C' is touched to 'B', charge flows to it until the two of them are equally charged. How much is that ? Well, just before they touch, 'C' has half of an original charge, and 'B' has a full one, so 1/4 of an original charge flows from 'B' to 'C', and then each of them has 3/4 of an original charge.

To review what we have now: 'A' has 1/2 of its original charge, and 'B' has 3/4 of it.

The force between any two charges is:

F = (a constant) x (one charge) x (the other one) / (the distance between them)².

For 'A' and 'B', the distance doesn't change, so we can leave that out of our formula.

The original force between them was 3 = (some constant) x (1 charge) x (1 charge).

The new force between them is F = (the same constant) x (1/2) x (3/4) .

Divide the first equation by the second one, and you have a proportion:

3 / F = 1 / ( 1/2 x 3/4 )

Cross-multiply this proportion:

3 (1/2 x 3/4) = F

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That's my story, and I'm sticking to it.

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3 years ago

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

$I = \dfrac{m(2gh - v^2)r^2}{v^2}$

$I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}$

$I =mr^2(0.287)$

I = 0.287 MR²

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