What happens to the temperature as altitude increases in the exosphere? Does it increase or decrease the higher it goes?

An earth satellite travels in a circular orbit at 20,000 mph if the radius of the orbit is 4,300 mi what angular velocity is gen

Bumek [7]

Answer:

0.00129rad/s

Explanation:

The angular velocity is expressed as;

v = wr

w is the angular velocity

r is the radius

Given

v = 20,000 mph

r = 4300mi

Get w;

w = v/r

w = 20000* 0.44704/4300*1609.34

w = 8940.8/6,920,162

w = 0.00129rad/s

Hence the angular velocity generated is 0.00129rad/s

5 0

3 years ago

The energy of the electron in Hydrogen atom can be shown to be given by En = -13.6/n^2 eV, where n represents the principal quan

Tpy6a [65]

Answer:

This represents radiation in ultra-violet region .

Explanation:

Energy of the orbit where n = 3 is given as follows

$E_3 =- \frac{13.6 }{3^2}$

$E_3 =- \frac{13.6 }{9}$ = -1.511 eV

Energy of the orbit where n = 1 is given as follows

$E_1 =- \frac{13.6 }{1^2}$

$E_1 =- \frac{13.6 }{1}$ = 13.6 eV

Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6

= 12.089 eV.

The wavelength of light having this energy in nm is given by the expression as follows

Wavelength in nm = 1244 / energy in eV

= 1244 / 12.089

= 102.90 nm

This represents radiation in ultra-violet region .

8 0

3 years ago

A car travelling at 15 m/s comes to rest in a distance of 14 m when the brakes are applied.

stira [4]

Answer:

-8.04 m/s2

Explanation:

To find the answer to this, you have to use the 4th kinematic equation:

$v^{2} = v^{2}_{0} + 2ax$

You plug into the equation to get:

$0 = 15^{2} + 2a(14)$

solve for a to get

-8.04 m/s2

3 0

3 years ago

Why would you expect the speed of light to be slightly less in the atmosphere then in a vacuum?

azamat

The speed of light to be slightly less in atmosphere then in vacuum because of absorption and re-emission of light by the atmospheric molecules occurred when light travels through a material

Explanation:

When light passes through atmosphere, it interacts or transmits through the transparent molecules in atmosphere. In this process of transmission through atmosphere, the light will be getting absorbed by them and some will get re-emitted or refracted depending upon wavelength.

But in vacuum the absence of any kind of particles will lead to no interaction and no energy loss, thus the speed of the light will be same in vacuum while due to interactions with molecules of atmosphere, there speed will be slightly less compared to in vacuum.

7 0

3 years ago

A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, $O=(0\ nm,0\ nm)$

Position of desired magnetic field, $D\equiv(1\ nm,8\ nm)$

Magnitude of desired magnetic field, $E=0\ N.C^{-1}$

Let q be the positive charge magnitude placed at origin.

We know the distance between the two Cartesian points is given as:

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

For the electric field effect to be zero at point D we need equal and opposite field at the point.

$\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }$

$\therefore (1-0)^2+(8-0)^2=r^2$

$r^2=65\ nm$

$r=\sqrt{65}$

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

$x=2\times 1=2\ nm$

and

$y=2\times 8=16\ nm$

3 0

3 years ago