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horsena [70]
2 years ago
6

If the bar magnet is flipped over and the south pole is brought near the hanging ball, the ball will be?

Physics
1 answer:
disa [49]2 years ago
7 0

The ball may attracted to the magnet.

<h3>How can we understand that the hanging ball will be attracted to the magnet or not?</h3>
  • From the question, we understand that the ball is attracted by the north pole of the bar magnet, then the bar magnet flipped over and the south pole is brought near the hanging ball.
  • As we know, in this type of experiments of bar magnet most of the times the ball is made out of steel.
  • Steel is a magnetic material.
  • Magnetic materials gets attracted  to the magnet at both the North and South pole.
  • This can be compared to how neutral objects also gets attracted to the positively and negatively charged rods through the Polarization force.

So, If the bar magnet is flipped over and the south pole is brought near the hanging ball, The ball will be attracted to the magnet.

Learn more about the bar magnet:

brainly.com/question/27943723

#SPJ4

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a_c = \frac{V^2}{R}

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a = \sqrt{a_t^2+a_r^2}

5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

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PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 6.75ft/s^2

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6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

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PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a_r = \frac{v^2}{R}

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Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

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a = 1.7187ft/s^2

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

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