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lorasvet [3.4K]
3 years ago
9

Help quick p;leaseeeeeeeeeeee its about direction and magnitude

Physics
1 answer:
KonstantinChe [14]3 years ago
7 0

Explanation:

The net force of each square is the combination of the forces in each direction. The direction is the... direction the square would go in due to the net force. The magnitude of the net force is how large it is. So if you had a force pushing 2N to the left and 1N to the right, then the net force would be 1N to the left; because the two oppose eachother. If they were going in the same direction, then they'd add to each other. And perpendicular net forces (like one pushing up and another pushing left) can create net forces in diagonal directions.

I'm not going to do all of these for you because they're basically all the same thing and it's good practice for you anyway. But I'll do the first three just so you can get the idea:

1. The net force's magnitude is 4N and it's direction is to the right.

2. The net force's magnitude is 4N and it's direction is to the left.

3. The net force's magnitude is 0N and it has no direction because they are equal forces acting in opposite directions.

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The amount of heat needed to increase the temperature of a substance by \Delta T is given by
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!! (Credits to The Psychics Classroom) !!
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A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
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Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

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I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

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