Physics, 02.07.2019 04:20 Brendah4962

Star a has a radius of 200 000 km and a surface temperature of 6 000 k. star b has a radius of 400 000 km and a surface temperature of 3 000 k. the emissivity of both stars is the same. what is the ratio of the rate of energy radiated by star a to that of star b?

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Answer from: sk9600930

Explanation:

In order to solve this problem, the Stefan-Boltzmann law will be useful. This law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":

$P=\sigma A T^{4}$ (1)

Where:

is the energy radiated by a blackbody radiator per second, per unit area (in Watts).

$\sigma=5.6703(10)^{-18}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

is the Surface of the body

is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (although the radiation of stars like our Sun is quite close). Therefore, we will use the Stefan-Boltzmann law for real radiator bodies:

$P=\sigma A \epsilon T^{4}$ (2)

Where $\epsilon$ is the star's emissivity

Knowing this, let's start with the

We have two stars where the emissivity $\epsilon$ of both is the same:

Star A with a radius $r_{A}=200000km$ and a surface temperature $T_{A}=6000K$ .

Star B with a radius $r_{B}=400000km$ and a surface temperature $T_{B}=3000K$ .

And we are asked to find the ratio of the rate of energy radiated by both stars:

$\frac{P_{A}}{P_{B}}$ (3)

Where $P_{A}$ is the rate of energy radiated by Star A and $P_{B}$ is the rate of energy radiated by Star B.

On the other hand, with the radius of each star we can calculate their surface area, using the formula for tha area of a sphere (assuming both stars have spherical shape):

$A_{A}=4\pi r_{A}^{2}$ (4)

$A_{B}=4\pi r_{B}^{2}$ (5)

Writting the Stefan-Boltzmann law for each star, taking into consideration their areas:

$P_{A}=\sigma (4\pi r_{A}^{2}) \epsilon {T_{A}}^{4}$ (6)

$P_{B}=\sigma (4\pi r_{B}^{2}) \epsilon {T_{b}}^{4}$ (7)

Substituting (6) and (7) in (3):

$\frac{\sigma (4\pi r_{A}^{2}) \epsilon {T_{A}}^{4}}{\sigma (4\pi r_{B}^{2}) \epsilon {T_{B}}^{4}}$ (8)

$\frac{P_{A}}{P_{B}}=\frac{{r_{A}}^{2} {T_{A}}^{4}}{{r_{B}}^{2} {T_{B}}^{4}}$ (9)

$\frac{P_{A}}{P_{B}}=\frac{{(200000km)}^{2} {(6000K)}^{4}}{{(400000km)}^{2} {(3000K)}^{4}}$ (10)

Finally:

$\frac{P_{A}}{P_{B}}=4$

Answer from: Quest

grav potential energy

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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