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astra-53 [7]
2 years ago
8

A flashlight contains a battery of two cells in series, with a bulb of resistance 12 Ohms. The internal resistance of each cell

is 0.26 ohms. If the potential difference across the bulb is 2.88V, what is the emf of EACH cell?
Physics
1 answer:
Elina [12.6K]2 years ago
4 0

Answer:

1.5024

Explanation:

Draw a diagram. Put the two cells in series. Now draw 3 resistors. Two of them equal 0.26 ohms each. The third one is the lightbulb which is 12 ohms.

R = 0.26 + 0.26 + 12 = 12.52

The bulb has a voltage of 2.88 volts across it. You can get the current from that.

i = E / R

i = 2.88 / 12 =

i = 0.24 amps.

Now you can get the voltage drop across the two cells.

E = ?

R = 0.26

i = 0.24 amps

E = 0.26 * 0.24

E = 0. 0624

Finally divide the 2.88 by 2 to get 1.44

Each cell has an emf of 1.44 + 0.0624 = 1.5024

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du
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With constant angular acceleration \alpha, the disk achieves an angular velocity \omega at time t according to

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For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

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Therefore on solving the above differential equation we get,

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Answer:

1968

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