Ask question

Ask question

All categories

2 years ago

5

what is the power of an electrical device which operates with a current of 12.4 A and a potential difference of 12 V

1 answer:

jek_recluse [69]2 years ago

3 0

148.8 Watts

Explanation:

P = VI

= (12 V)(12.4 A)

= 148.8 Watts

You might be interested in

An electron moving in the direction of the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in

Gnom [1K]

Answer:

<em>-z axis</em>

Explanation:

According to the left hand rule for an electron in a magnetic field, hold the thumb of the left hand at a right angle to the rest of the fingers, and the rest of the fingers parallel to one another. If the thumb represents the motion of the electron, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the electron. In this case, the left hand will be held out with the thumb pointing to the right (+x axis), and the palm facing your body (-y axis). The magnetic field indicated by the other fingers will point down in the the -z axis.

5 0

2 years ago

A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb

densk [106]

Voltage=V=120V
Power=P=60W

Current=I

$\\ \rm\rightarrowtail I=\dfrac{P}{V}=\dfrac{60}{120}=\dfrac{1}{2}=0.5A$

8 0

2 years ago

The mass of the jupiter is 19*10^26kg &the mass of the earth is 6*10^24kg.if the distance between the jupiter &earth is

Firlakuza [10]

Answer:
The Gravitational Force between the 2 masses is approximately 1.209ｘ10^32 Newton’s

Explanation:

6 0

2 years ago

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

or

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.

6 0

2 years ago

Please help!<br> Drag and drop the words into the correct locations

Stells [14]

Answer:

background radiation

uniform

13700000000 years ago

7 0

2 years ago

Read 2 more answers