All categories

3 years ago

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]3 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

You might be interested in

Which of the following is an example of heat convection?

lora16 [44]

I believe it’s a liquid inside a beaker on a hot Bunsen burner (c)

This is because : Everyday Examples of Convection
Boiling water - The heat passes from the burner into the pot, heating the water at the bottom. Then, this hot water rises and cooler water moves down to replace it, causing a circular motion. Radiator - Puts warm air out at the top and draws in cooler air at the bottom.

Not sure if it’s right tho!

8 0

3 years ago

Read 2 more answers

Witch planet takes 84 earth year's to orbit the sun just once

alekssr [168]

Uranus takes 84 earth years to make a full rotation around the sun<span />

8 0

3 years ago

Read 2 more answers

A 4-kg cat is resting on a bookshelf that is 2 m high. What is the cats gravitational potential energy relative to the floor if

ra1l [238]

The most exact answer is 78.4J also in this kind of options we can say answer "d"

3 0

3 years ago

Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher

siniylev [52]

Answer:

$v = 4.44 \times 10^5 m/s$

Explanation:

By Einstein's Equation of photoelectric effect we know that

$h\nu = W + \frac{1}{2}mv^2$

here we know that

$h\nu$ = energy of the photons incident on the metal

$W$ = minimum energy required to remove photons from metal

$\frac{1}{2}mv^2$ = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

$W = \frac{hc}{351 nm}$

here we know that

$hc = 1242 eV-nm$

now we have

$W = \frac{1242}{351} = 3.54 eV$

now by energy equation above when photon of 303 nm incident on the surface

$\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2$

$4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2$

$(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2$

$8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2$

$v = 4.44 \times 10^5 m/s$

6 0

3 years ago

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe

Allushta [10]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

$P = \sigma Ae\Delta T^4$