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MAVERICK [17]
4 years ago
9

When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in f

lux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The braking force on the magnet is nearly equal to its weight, so it falls very slowly. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the magnet is pushed, forcefully, toward the block, the rate of change of flux is much higher than this. When the magnet is moving much more quickly than it will fall unaided, what is the direction of the net force on the magnet?
Physics
1 answer:
Pepsi [2]4 years ago
7 0

Answer:

<em>The net force is directed downwards</em>.

Explanation:

Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.

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Which statement is correct? (2 points)
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When a neutral atom looses an electron to another neutral atom, two charged atoms are created.

Explanation:

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3 years ago
An FM radio consists of a series RLC circuit with a 6.57 pF (peco – 10-12) capacitor. If a person is dialed into a station broad
faust18 [17]

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3.6 x 10⁻⁷ H

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4 0
3 years ago
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Leni [432]

Answer: 12.0 m/s^2

Explanation:

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\tau = W \times OM\\ or\\ \tau = mg\times \left(\dfrac{L}{2}\right)\sin 55^\circ ....(i)

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

\tau = I_{rod}\ \alpha......(ii)

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mg\times \left(\frac{L}{2}\right)\sin 55^\circ = \left(\frac{mL^2}{3}\right)\alpha\ \ \ \ \ \ \ (\because I_{rod} = \dfrac{mL^2}{3}) \\ \\ \alpha = \left(\frac{3g}{2L}\right)\sin 55^\circ\\ \\ \alpha = \left(\frac{3\times 9.8}{2\times 2.4}\right)\sin 55^\circ\\ \\ \boxed{\alpha = 5.02\ rad/s^2} }

The acceleration of the end of the rod farthest from the link is given by:

a = L\alpha\\ \\ a = (2.4\ m)(5.02\ rad/s^2)\\ \\ \boxed{a = 12.0\ m/s^2}

7 0
2 years ago
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