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Ann [662]
3 years ago
13

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 310 turns. the inner solenoid is 25.0 cm long

and has a diameter of 2.00 cm. at a certain time, the current in the inner solenoid is 0.116 a and is increasing at a rate of 1.85 103 a/s. (a) for this time, calculate the average magnetic flux through each turn of the inner solenoid. wb (b) for this time, calculate the mutual inductance of the two solenoids. h (c) for this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid. v

Physics
1 answer:
Ymorist [56]3 years ago
6 0
(a) Image 1
(b) Image 2
(c) Image 3

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An eccentric inventor attempts to levitate a cork ball by wrapping it with foil and placing a large negative charge on the ball
djverab [1.8K]

Answer:

because of the idea that like charges get repulsion as a force.

Explanation:

because you wrap the ball with foil, the negative charges will leave the foil and go into the ball by induction. This leaves the foil as a positively charged particle since its electrons left it for the ball, making the ball a negatively charged particle. but if you bring the negative charge near the foil, the electrons will transfer from that and go into the foil, making it negatively charged. Now, because the ball and the foil have the same charge, they repel. the foil flies off.

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2 years ago
PLEASE HELP!!!
pickupchik [31]
The answer is ultraviolet rays

7 0
3 years ago
Viewing moving objects from pond water under his microscope and named then animalcules
pogonyaev
I think it was either Robert Hook or Anton Van Leevwen Hoek
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3 years ago
Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart.
serious [3.7K]

Answer:

1.6\times 10^{-7} N

2.4\times 10^{-7} N

Explanation:

i_{1} = 1 A

i_{2} = 4 A

r = distance between the two wire = 5 m

F = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

F = \frac{\mu _{o}}{4\pi }\frac{2i_{1}i_{2}}{r}

F = (10^{-7})\frac{2(1)(4)}{5}

F = 1.6\times 10^{-7} N

r'} = distance of each wire from the midpoint = 2.5 m

Magnetic field midway between the two wires is given as

B = \frac{\mu _{o}}{4\pi } \left \left ( \frac{2i_{2}}{r'} \right - \frac{2i_{1}}{r'} \right \right ))

B = (10^{-7}) \left \left ( \frac{2(4)}{2.5} \right - \frac{2(1)}{2.5} \right \right ))

B = 2.4\times 10^{-7}

5 0
3 years ago
If the cross-section of a wire of fixed length is doubled, how does the resistance of that wire change? (this is for studying fo
Ymorist [56]
If the cross-section of a wire of fixed length is doubled,  the resistance of that wire change into doubled.We know that <span>the total </span>length<span> of the wires will </span>affect<span> the amount of </span>resistance. <span> The longer the wire, the more </span>resistance<span> that there will be so the answer is doubled.</span>
6 0
3 years ago
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