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Elanso [62]
3 years ago
5

If a magnet repels a metal object what can you say about the pole of the magnet and the pole of the object?

Physics
2 answers:
sleet_krkn [62]3 years ago
7 0
Poles are the same D is the right answer
Ilya [14]3 years ago
3 0

Answer:

The correct answer is option D. The poles are the same.

Explanation:

Hello!

Let's solve this!

For an object to be attracted to a magnet, the charges must be opposite. But when, otherwise, they repel, it means that the charges (the poles) are the same.

In this case we conclude that the correct answer is option D. The poles are the same.

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Wich advantage of reproduction does the graph shown
inn [45]

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No photo or graph is there

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2 years ago
Joe and Max shake hands and say goodbye. Joe walks east 0.50 km to a coffee shop, and Max flags a cab and rides north 3.45 km to
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Answer:

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8 0
3 years ago
The maximum tension the wire can withstand without breaking is 300 N . A 0.800 kg projectile traveling horizontally hits and emb
topjm [15]

Answer:

Let's investigate the case where the cable breaks.

Conservation of angular momentum can be used to find the speed.

\vec{L}_1 = \vec{L}_2\\\vec{L}_1 = m\vec{v_0} \\\vec{L}_2 = I\vec{\omega}\\

The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>

I = I_{projectile} + I_{ball}\\I = mr^2 + mr^2\\I = 2mr^2

where r is the length of the cable.

<u>After the collision, the ball and the projectile makes a circular motion because of the cable.</u> So, the force (tension) in circular motion is

F = \frac{mv^2}{r}

The relation between linear velocity and the angular velocity is

v = \omega r

So,

F = \frac{m(\omega r)^2}{r} = m\omega^2 r = 300\\mv_0r  =I\omega\\\\\omega = mv_0r/I\\300 = m(\frac{mv_0r}{I})^2r = m(\frac{mv_0r}{2mr^2})^2r = m(\frac{v_0}{2r})^2r = \frac{mv_0^2r}{4r^2} = \frac{mv_0^2}{4r}\\300 = \frac{0.8v_0^2}{4r}\\1500 = v_0^2/r\\v_0 = \sqrt{1500r}

As can be seen, the maximum velocity for the projectile without breaking the cable is \sqrt{1500r}, where r is the length of the cable.

6 0
3 years ago
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