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3 years ago

What can happen to an eletron in an atom when thr atom gains or loses energy

Physics

1 answer:

RideAnS [48]3 years ago

3 0

It moves closer or farther to the protons and neutrons. Hope this helps.

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5. Kelvin applied a rightward force of 302 N to a 29 kg desk to accelerate it across the floor. The

Naddik [55]

The force of friction is 213.2 N

Explanation:

The frictional force acting on an object sliding on a surface is given by:

$F_f = \mu mg$

where:

$\mu$ is the coefficient of friction

m is the mass of the object

g is the acceleration of gravity

In this problem we have

$\mu=0.750$

m = 29 kg

$g=9.8 m/s^2$

Therefore, the force of friction is

$F_f = (0.750)(29)(9.8)=213.2 N$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

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#LearnwithBrainly

3 0

3 years ago

One of the foci for the moon's orbit would be the

suter [353]

The question is oversimplified, and pretty sloppy.

Relative to the Earth . . .
The Moon is in an elliptical orbit around us, with a period of
27.32... days, and with the Earth at one focus of the ellipse.

Relative to the Sun . . .
The Moon is in an elliptical orbit around the Sun, with a period
of 365.24... days, and with the Sun at one focus of the ellipse,
and the Moon itself makes little dimples or squiggles in its orbit
on account of the gravitational influence of the nearby Earth.

I'm sorry if that seems complicated. You know that motion is
always relative to something, and the solar system is not simple.

5 0

2 years ago

Read 2 more answers

What is the atmospheric pressure 1.00 km above the surface of Venus? Express your answer in Earth-atmospheres.

EleoNora [17]

Below is an attachment containing the solution.

4 0

3 years ago

An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mV/m that points in

weqwewe [10]

Answer:

a) $t=1.4\times 10^{-5}\ s$

b)S= 46.4 cm

Explanation:

Given that

Velocity = 16 Km/s

V= 16,000 m/s

E= 27 mV/m

E=0.027 V/m

d= 22.5 cm

d= 0.225 m

lets time taken by electron is t

d = V x t

0.225 = 16,000 t

$t=1.4\times 10^{-5}\ s$

We know that

F = m a = E q ------------1

Mass of electron ,m

$m=9.1\times 10^{-31}\ kg$

Charge on electron

$q=1.6\times 10^{-19}\ C$

So now by putting the values in equation 1

$a=\dfrac{E q}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 0.027}{9.1\times 10^{-31}}\ m/s^2$

$a=4.74\times 10^{9}\ m/s^2$

$S= ut+\dfrac{1}{2}at^2$

Here initial velocity u= 0 m/s

$S= \dfrac{1}{2}\times 4.74\times 10^{9}\times (1.4\times 10^{-5})^2\ m$

S=0.464 m

S= 46.4 cm

S is the deflection of electron.

4 0

3 years ago

Read 2 more answers

An astronaut having mass 320 kg with equipment included is attempting an untethered space walk. The astronaut is initially at re

ExtremeBDS [4]

This can be solved using momentum balance, since momentum is conserved, the momentum at point 1 is equal to the momentum of point 2. momentum = mass x velocity
m1v1 = m2v2
(0.03kg x 900 m/s ) = 320(v2)
v2 = 27 / 320
v2 = 0.084 m/s is the speed of the astronaut

7 0

3 years ago