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Ratling [72]
4 years ago
6

A circular coil lies flat on a horizontal surface. A bar magnet is held fixed above the center of the coil with its north pole p

ointing downward. What is the direction of the induced current in the coil
Physics
1 answer:
Alex17521 [72]4 years ago
5 0

Answer:

Explanation:

If the magnet is dropped from this position, then the direction of the induced current in the coil would be anticlockwise. Lenz's law allows us to predict the direction of the induced current in a circuit. It is also known as the Right hand rule. It states that the current that is being induced in a particular circuit is as a result of the change in the magnetic field in order to oppose the change in the magnetic flux and also to apply an mechanical force that is opposing the motion. Because of this, the current is pointed to the direction of  the pole, to oppose this the induced current should be directed anticlockwise.

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Converting 16.2 pounds to 1 newton
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16.2 pounds doesn't convert to 1 newton.
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3 years ago
A military helicopter on a training mission is flying horizontally at a speed of 90.0 m/s when it accidentally drops a bomb (for
Elena-2011 [213]

Answer:

1) 10.1 s  2) 909 m 3) 90.0 m/s 4) -99m/s 5) just over the bomb.

Explanation:

1)

  • In the vertical direction, as the bomb is dropped, its initial velocity is 0.
  • So, we can find the time required for the bomb to reach the earth, applying the following kinematic equation for displacement:

       \Delta y = \frac{1}{2}*a*t^{2} (1)

  • where Δy = -500 m (taking the upward direction as positive).
  • a=-g=-9.8 m/s²
  • Replacing these values in (1), and solving for t, we have:

       t =\sqrt{\frac{2*\Delta y}{-g}} = \sqrt{\frac{2*(-500m)}{-9.8m/s2}} = 10.1 s

  • The time required for the bomb to reach the earth is 10.1 s.

2)

  • In the horizontal direction, once released from the helicopter, no external influence acts on the bomb, so it will continue moving forward at the same speed. that it had, equal to the helicopter.
  • As the time must be the same for both movements, we can find the horizontal displacement just as the product of this speed times the time, as follows:

       x = v_{0x} * t = 90.0 m/s * 10.1 s = 909 m.

3)

  • The horizontal component of the bomb's velocity is the same that it had when left the helicopter. i.e. 90 m/s.

4)

  • In order to find the vertical component of the bomb's velocity just before it strikes the earth, we can apply the definition of acceleration, remembering that v₀ = 0, as follows:

        v_{f} = -g*t = -9.8 m/s2*10.1 s = -99 m/s

5)

  • If the helicopter keeps flying horizontally at the same speed, it will be always over the bomb, as both travel horizontally at the same speed.
  • So, when the bomb hits the ground, the helicopter will be exactly over it.

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