Placing the north pole of a magnet near the south pole of another magnet results in:

Physics

1 answer:

MakcuM [25]3 years ago

8 0

Answer:

An attractive force between the magnets. A is the correct answer if it was the north pole facing another north pole magnet the answer would have been a repulsive force between the magnets because the north pole of a magnet does not attract instead it separates from each other which is a repulsive force the best answer is A.

Explanation:

the southern pole of a magnet is a positive force while the northern pole of a magnet is negative .

Note: Magnet can attract each other only when they are facing the opposite directions a north pole will attract a south pole; the magnets pull on each other. But the two north poles will push each other away.

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Suppose a star the size of our Sun, but with mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 7.0 days.

kvasek [131]

Answer:

Its rotation will be 3.89x10⁴ rad/s.

Explanation:

We can find the rotation speed by conservation of the angular momentum:

$L_{i} = L_{f}$

$I_{i}\omega_{i} = I_{f}\omega_{f}$ (1)

The initial angular speed is:

$\omega_{i} = \frac{1 rev}{7 d} = 0.14 \frac{rev}{d}$

The moment of inertia (I) of a sphere is:

$I = \frac{2}{5}mr^{2}$ (2)

Where m is 9 times the sun's mass and r is the sun's radius

By entering equation (2) into (1) we have:

$\frac{2}{5}m_{i}r_{i}^{2}\omega_{i} = \frac{2}{5}m_{f}r_{f}^{2}\omega_{f}$

$9m_{sun}(696342 km)^{2}0.14\frac{rev}{d} = \frac{3}{4}9m_{sun}(13 km)^{2}\omega_{f}$

$\omega_{f} = \frac{4}{3}*0.14 \frac{rev}{d}(\frac{696342 km}{13 km})^{2} = 5.36 \cdot 10^{8} \frac{rev}{d}*\frac{1 d}{24 h}*\frac{1 h}{3600 s}*\frac{2\pi rad}{1 rev} = 3.89 \cdot 10^{4} rad/s$