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Vsevolod [243]
2 years ago
15

Question 14:A light bulb is connected to a 110-V source. What is the resistance of this bulb if it is a 100-W bulb

Physics
1 answer:
miss Akunina [59]2 years ago
6 0

By the use of Ohm's law, the resistance of the bulb is 121 ohms

<h3>What is electrical power?</h3>

Let us recall that power is the rate of doing work. Now we know that the voltage of the power source is 110-V and the power is 100-W.

Since;

P= IV

I = P/V

I = 100/110

I = 0.91 A

Again;

V= IR

R = V/I

R = 110/0.91

R = 121 ohms

Learn more about resistance: brainly.com/question/11431009

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Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t
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1.9161676647\times 10^{13}\ m/s^2 or 1.9532799844\times 10^{12}g

23.4843749996 m

Yes

Explanation:

E = Electric field = 2\times 10^5\ N/C

c = Speed of light = 3\times 10^8\ m/s

m = Mass of proton= 1.67\times 10^{-27}\ kg

q = Charge of electron = 1.6\times 10^{-19}\ C

Acceleration is given by

a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g

The acceleration is 1.9161676647\times 10^{13}\ m/s^2 or 1.9532799844\times 10^{12}g

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

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