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2 years ago

Write down short notes on:a. Electrical energy

Physics

1 answer:

PilotLPTM [1.2K]2 years ago

4 0

Answer:

Explanation:

Electrical energy is energy derived from electric potential energy or kinetic energy.

Or,

Electrical energy is a form of energy resulting from the flow of electric charge. Lightning, batteries and even electric eels are examples of electrical energy.People use electricity for lighting, heating, cooling, and refrigeration and for operating appliances, computers, electronics, machinery, and public transportation systems.

Hope it helped you.

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Which property of metals allows corrosion to take place? reactivity malleability ductility conductivity Description

horrorfan [7]

Your best bet would be reactivity

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2 years ago

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It cannot be constant because if it does not change and each time it increases its strength and speed.

Explanation:

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2 years ago

A force F acts in the forward direction on a cart of mass m. A friction force ff opposes this motion. Part A Use Newton's second

stellarik [79]

Answer:

a = F-ff/m

Explanation:

According to Newton's second law of motion which states that "the rate of change in momentum of a body is directly proportional to the applied force F and acts in the direction of the force.

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F = ma

Since two forces acts on the cart i.e the moving force F and the frictional force Ff , we will take the sum of the forces.

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F - ff = ma

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3 years ago

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$ (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

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3 years ago

Use the exact values you enter to make later calculations. You repeat the same experiment that you did in the lab with the force

frutty [35]

Answer:

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3 years ago

Write down short notes on:a. Electrical energy​

Write down short notes on:a. Electrical energy