All categories

3 years ago

In which region of the electromagnetic spectrum is most of earth's outgoing terrestrial radiation?

1 answer:

6 0

The answer is Infrared. The infrared of the electromagnetic spectrum is most of earth's outgoing terrestrial radiation. <span>Earth is the hot body with temperature of 30 degrees on the average.</span>

You might be interested in

What formula should I use?

Elodia [21]

These nuts on your chin

7 0

3 years ago

A star ending its life with a mass of four to eight times the Sun's mass is expected to collapse and then undergo a supernova ev

Zielflug [23.3K]

Answer:

r = 16 Km

Explanation:

given

m_n= 1.67 x 10^-27 Kg

M_star = 3.88 x 10^30 Kg

A= M_star/m_n

A= 3.88*10^30/1.67 x 10^-27

A=2.28 *10^57 neutrons A = The number of neutrons

we use the number of neutrons as a mass number because the star has only neutrons. = 1.2 x 10-15 m

r = r_o*A^1/3

r = 1.2*10^-15*2.28 *10^57^1/3

r = 16 Km

8 0

3 years ago

Which shows the formula for converting from kelvins to degrees Celsius? °C = (9/5 × K) +32 °C = 5/9 × (K – 32) °C = K – 273 °C =

madam [21]

The freezing point of the water is 0 C , and it equals to 273 K

Then, To convert from Kelvins degrees to Celsius degrees we use the relation

$K = C + 273$

Also,

$C = K - 273$

5 0

3 years ago

Read 2 more answers

A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p

alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

$F=qvB$

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

$qvB = \frac{mv^2}{r}$

which can be rewritten as

$v=\frac{qB}{mr}$

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

$\frac{2\pi r}{T}=\frac{qB}{mr}$

So, we get:

$T=\frac{2\pi m r^2}{qB}$

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) $a. f/10$

The frequency of revolution of a particle in uniform circular motion is

$f=\frac{1}{T}$

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

$f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}$

6 0

3 years ago

A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a rope that goes over an ideal pulley. If the masses are

irga5000 [103]

Answer:

The time taken will be 0.553 seconds.

Explanation:

We should start off by finding the force exerted by the rope on the 3kg weight in this case.

Weight of 5kg mass = 5 * 9.81 = 49.05 N

Weight of 3kg mass = 3 * 9.81 = 29.43 N

The force acting upward on the 3kg mass will equal the weight of the 5kg mass. Thus the resultant force acting on the 3kg mass is:

Total force = 49.05 - 29.43 = 19.62 N (upwards)

We can now find the acceleration:

F = m * a

19.62 = 3 * a

a = 6.54 m/s^2

We now use the following equation of motion to get the time taken to travel 1 meter:

$s=u*t+\frac{1}{2} (a*t^2)$

$1=0*t+\frac{1}{2} (6.54*t^2)$

t = 0.553 seconds

4 0

3 years ago