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3 years ago

When a charge is discharged to the earth, we call it?

Physics

1 answer:

Tems11 [23]3 years ago

3 0

Answer:

lightning

Explanation:

correct answer is Lightning because it is the naturally occurring electrostatic discharge, during 2 electrically charge field in atmosphere and on the ground temporarily equalize themselves and they instantly release energy.
As it is released cloud so we called it lightning but a spark is generated when the electric field strength is sufficient to ionize the molecules of air, creating a small conduction path through the air.
This is usually a cascade process, because free electrons accelerate in the electric field and ionize more molecules by colliding.

You might be interested in

Which of the following statements are true?

Tems11 [23]

Answer:

1) A time-varying magnetic field will produce an electric field.

4) A time-varying electric field will produce a magnetic field.

Explanation:

1) A time-varying magnetic field will produce an electric field.

TRUE

time varying magnetic field will produce electric field which is given as

$E = \frac{r}{2}\frac{dB}{dt}$

2) Time-varying electric and magnetic fields can propagate through space only if there is no matter in their path.

FALSE

Time varying electric field and magnetic field will induce each other and it can travel through any medium as well as it can travel without any medium also

3) Electric and magnetic fields can be treated independently only if they vary in time.

FALSE

electric field can be due to stationary charge and magnetic field due to current carrying elements so it is not compulsory to have time varying

4) A time-varying electric field will produce a magnetic field.

TRUE

Time varying electric field will produce magnetic field given as

$B = \frac{\mu_o\epsilon_o A}{2r}\frac{dE}{dt}$

3 0

3 years ago

The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proto

igor_vitrenko [27]

a) The angular velocity of the electron is $4.12\cdot 10^{16} rad/s$

b) The number of revolutions per second is $6.54\cdot 10^{15}$ rev/s

c) The centripetal acceleration of the electron is $8.98\cdot 10^{22} m/s^2$

Explanation:

This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.

The angular velocity of an object in uniform circular motion is given by

$\omega = \frac{v}{r}$

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

$v=2.18 \cdot 10^6 m/s$

$r=5.29\cdot 10^{-11} m$

Therefore, the angular velocity is

$\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s$

The period of revolution of the electron is given by

$T=\frac{2\pi}{\omega}$

where

$\omega = 4.12\cdot 10^{16}rad/s$ is the angular velocity

Substituting,

$T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s$

The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:

$f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15}$ rev/s

The centripetal acceleration of an object in circular motion is

$a=\frac{v^2}{r}$

where

v is the linear speed

r is the radius of the circle

For the electron, we have

$v=2.18 \cdot 10^6 m/s$

$r=5.29\cdot 10^{-11} m$

Therefore, the centripetal acceleration is

$a=\frac{(2.18\cdot 10^6)^2}{5.29\cdot 10^{-11}}=8.98\cdot 10^{22} m/s^2$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0

4 years ago

PLEASE HELP!!! When you lift a book from the ground to your desk, what kind of work do you do, negative or positive? By lifting

Snowcat [4.5K]

When someone lifts a book from the ground, the work you use is positive. By lifting the book, you change it's energy and it's original place The book gains, kinectic energy.

Hope I helped.

8 0

3 years ago

Approximately, What is the value of the Hubble Constant, as measured by scientists? Hypothetically, if the value of the Hubble C

Serhud [2]

Answer:

The current value of the Hubble's constant = 73 km/sec/Mpc.

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

Explanation:

The current value of the Hubble's constant = 73 km/sec/Mpc. However, recent discoveries in the cosmology contradicts the idea of Hubble constant as being fixed. Some scientists are not agreeing on this value and the debate is going on.

Hubble law states that how fast universe is expanding or in other words, galaxies are expanding separating with a speed directly proportional to the distance of galaxies to the earth.

Hence,

v is directly proportional to d

where, v = apparent velocity

d = distance

if we equate velocity and distance then there comes Hubble constant.

v = $H_{0}$ x d

$H_{0}$ = 73 km/sec/Mpc

where, Mpc = Mega Parsec = 1 Mpc = 3.086 x $10^{19}$ km

We can use Hubble constant to tell the age of universe.

t = d/v

t = d/( $H_{0}$ xd)

t = 1/ $H_{0}$

Scientist calculated the age of universe by using Hubble constant, which is 13.4 billion years.

Now, if we hypothetically change the value of Hubble constant,

from $H_{0}$ = 73 km/sec/Mpc to $H_{0}$ = 700 km/sec/Mpc

then the age of universe will be:

t = 1/ $H_{0}$

first convert the units of new $H_{0}$ into 1/s

$H_{0}$ = (700) x (/3.08 x $10^{19}$ )

$H_{0}$ = 227.27 x $10^{-19}$ = 2.27 x $10^{-21}$ 1/s

So,

Age of universe will be:

t = 1/ $H_{0}$ = 1/2.27x $10^{-21}$ 1/s

t = 2.27 x $10^{21}$ s

t = 71.9 trillion years

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

6 0

3 years ago

A man of mass 81 kg escapes from a burning building by jumping from a window situated 32 m above a catching net. The acceleratio

Dafna1 [17]

Answer:

The speed of man before he hits the ground is <u>23.35 m/s</u>

Explanation:

We know that:

Weight of Man - Force of Friction = Unbalanced Force

but, from Newton's 2nd Law of Motion:

unbalanced force = ma

Therefore,

W - F = ma

a = (W - F)/m

a = (mg - F)/m

where,

m = 81 kg

g = 9.8 m/s²

F = 103 N

a = [(81 kg)(9.8 m/s²) - 103 N]/81 kg

a = 8.52 m/s²

using 3rd equation of motion:

Vf² - Vi² = 2ah

here,

Vi = initial velocity = 0 m/s

Vf = Final Velocity before he hits ground = ?

Vf² - 0² = 2(8.52 m/s²)(32 m)

Vf = √545.28 m²/s²

4 0

3 years ago