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Leno4ka [110]
3 years ago
15

The operating potential difference of a light bulb is 120 V. The power rating of the bulb is 70 W. (a) Find the current in the b

ulb. A (b) Find the bulb's resistance.
Physics
1 answer:
3241004551 [841]3 years ago
6 0

Answer: I= 0.583A

R=205.83ohms

Explanation: Power= I*V

Where I= current and V = potential difference

I=P/V

I= 70/120

I=0.583A

To find resistance

Using ohms law

V= IR

R=V/I

R= 120/0.583

R= 205.83ohms

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Answer:

2.09\ \text{m/s}

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As the momentum is conserved in the system we have

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As energy in the system is conserved we have

K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}

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Answer:

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