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madreJ [45]
3 years ago
11

Suppose that when you move the north pole of a bar magnetic toward a coil it induces a positive current in the coil. The strengt

h of the field produced by an electromagnetic can be controlled electronically. Suppose you place a coil near the north pole of an electromagnet and increase the field while keeping everything stationary. Which one of the following will happen? a) A positive current will be induced in the coilb) A negative current will be induced in the coil c) No current will be induced in the coil since there is no relative motion.
Physics
1 answer:
dolphi86 [110]3 years ago
3 0

Answer:

<em>a) A positive current will be induced in the coil</em>

Explanation:

Electromagnetic induction is the induction of an electric field on a conductor due to a changing magnetic field flux. The change in the flux can be by moving the magnet relative to the conductor, or by changing the intensity of the magnetic field of the magnet. In the case of this electromagnets<em>, the gradual increase in the the electromagnet's field strength will cause a flux change, which will in turn induce an electric current on the coil.</em>

According to Lenz law, the induced current acts in such a way as to negate the motion or action that is producing it. <em>A positive current will be induced on the coil so as to repel any form of attraction between the north pole of the electromagnet and the coil</em>. This law obeys the law of conservation of energy, since work has to be done to move the move them closer to themselves.

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Furkat [3]
The answer is A because it describes both toy cars
6 0
3 years ago
A negative charge of -0.550 m exerts an upward <a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="5868766e
almond37 [142]

Answer:

a. +10.9μC

b. 0.600N and downward

Explanation:

To determine the magnitude of the charge, we use the force rule that exist between two charges which us expressed as

F=(kq₁q₂)/r²

since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as

q₂=Fr²/kq₁

q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)

q₂=0.054/4950

q₂=1.09*10⁻⁵c

q₂=10.9μC.

Hence the second charge is +10.9μC

b. From the rule of charges which state that like charges repel and unlike charges attract, we can conclude that the two above charges will attract since they are unlike charges. Hence the direction of the force will be downward into the second charge and the magnitude of the force will remain the same as 0.600N

8 0
3 years ago
If we have 300 centimeters,we would have
Leno4ka [110]
3 meters................
3 0
3 years ago
A square sheet of rubber has sided that are 20 cm long. If you stretch one side so that it is twice as long. What is the new are
Anika [276]

Answer:ignore this it was wrong

Explanation:

3 0
4 years ago
Sobre un barco, que se mueve en forma rectilínea, y con velocidad constante de 30 [km/h], se mueve un perro en el mismo sentido
almond37 [142]

Answer:

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

Explanation:

Para determinar la velocidad del perro con respecto al observador sentado desde la playa a través del concepto de velocidad relativa, descrito en la siguiente fórmula:

v_{P/B} = v_{P} - v_{B} (1)

Donde:

v_{P/B} - Velocidad del perro relativo al barco, en kilómetros por hora.

v_{P} - Velocidad del perro con respecto al observador, en kilómetros por hora.

v_{B} - Velocidad del barco con respecto al observador, en kilómetros por hora.

Si sabemos que v_{B} = 30\,\frac{km}{h} y v_{P/B} = 10\,\frac{km}{h}, entonces la velocidad del perro con respecto al observador es:

v_{P} = v_{B} + v_{P/B}

v_{P} = 30\,\frac{km}{h} + 10\,\frac{km}{h}

v_{P} = 40\,\frac{km}{h}

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

6 0
3 years ago
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