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larisa [96]
3 years ago
13

If R1 and R2 are in parallel and R3 is in series with them then equivalent resistance will be

Physics
1 answer:
pochemuha3 years ago
5 0

Answer:

Refer to the attachment!~

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A 9-volt battery is used to power two lightbulbs in a series circuit. The first light bulb has a
butalik [34]

Explanation:

Total Resistance: 3+2 = 5 ohms

V=9V

Power = V²/R

=9²/5

=81/5

=16.2 W

6 0
2 years ago
A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b
Natasha_Volkova [10]

Answer:

Part a)

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part b)

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

\int E. dA = \frac{q}{\epsilon_0}

E. 2\pi rL = \frac{\lambda L}{\epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part b)

Outside the outer cylinder we will again use Guass law

\int E. dA = \frac{q}{\epsilon_0}

E. 2\pi rL = \frac{\lambda L}{\epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

4 0
3 years ago
Can someone please help?
kenny6666 [7]

Answer:

uh with what?

Explanation:

4 0
2 years ago
When an electron is displaced in a semiconductor, the hole that's left behind is
morpeh [17]

Answer:

A. attracted to the negative terminal of the voltage source.

Explanation:

When an electron is displaced in a semiconductor, the hole that's left behind is

A. attracted to the negative terminal of the voltage source.

The electron leaving leaves a net + charge, which is attracted to the negative terminal.

8 0
2 years ago
Is this right at all
Alja [10]
It is correct! good job :)
7 0
3 years ago
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