3 years ago

What is the definition of electrical discharge

Physics

1 answer:

koban [17]3 years ago

7 0

Answer:

electrical conduction through a gas in an applied electric field

Explanation:

You might be interested in

You are playing catch with a friend in a moving train. When you toss the ball in the direction the train is moving, how does the

S_A_V [24]

I think to an observer outside the train, the speed of the ball will actually look like more than the speed of the train.

4 0

3 years ago

The de Broglie wavelength of an electron traveling at 7.0 x 107m/s is 1.0 x 10-11m. (Remember that me = 9.1 x 10-31kg and h = 6.

ivolga24 [154]

The De Broglie's wavelength of a particle is given by:

$\lambda=\frac{h}{p}$

where

$h=6.6 \cdot 10^{-34} Js$ is the Planck constant

p is the momentum of the particle

In this problem, the momentum of the electron is equal to the product between its mass and its speed:

$p=m_e v=(9.1 \cdot 10^{-31} kg)(7.0 \cdot 10^7 m/s)=6.4 \cdot 10^{-23} kg m/s$

and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:

$\lambda=\frac{h}{p}=\frac{6.6 \cdot 10^{-34} Js}{6.4 \cdot 10^{-23} kg m/s}=1.0 \cdot 10^{-11} m$

So, the answer is True.

7 0

3 years ago

Lithium is more active than aluminium A.True B.false

WITCHER [35]

A:True because Lithium is the lighest solid and metal and the third lightest element.

5 0

3 years ago

Read 2 more answers

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)