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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:

18 N/C

Explanation:

Given that:

Electric field constant, k = 9*10^9 N/c

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E = (18 * 10^9 * 10^-33) ÷ 10^-24

E = [18 * 10^(9-33)] ÷ 10^-24

E = (18 * 10^-24) / 10^-24

E = 18 * 10^-24+24

E = 18 * 10^0

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3 years ago
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To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

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\tau = (8i+6j)\times(-3i+4j)

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Therefore the torque on the particle about the origen is 50k

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cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}

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cos\theta = -0.24

\theta = cos^{-1} (-0.24)

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Therefore the angle between the ratio and the force is 103.88°

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