Which is the correct answer?

consider two stars, star a and star b. star a has a temperature of 4900 k , and star b has a temperature of 9900 k . how many ti

ValentinkaMS [17]

The Energy flux from Star B is 16 times of the energy flux from Star A.

We have Two stars - A and B with 4900 k and 9900 k surface temperatures.

We have to determine how many times larger is the energy flux from Star B compared to the energy flux from Star A.

<h3>State Stephen's Law?</h3>

Stephens law states that if E is the energy radiated away from the star in the form of electromagnetic radiation, T is the surface temperature of the star, and σ is a constant known as the Stephan-Boltzmann constant then-

$$\frac{Energy}{Area} = \sigma\times T^{4}$

Now -

Energy emitted per unit surface area of Star is called Energy flux. Let us denote it by E. Then -

$$E= \sigma\times T^{4}$

Now -

For Star A →

$T_{A}$ = 4900 K

For Star B →

$T_{B}$ = 9900 K

Therefore -

$$\frac{T_{B} }{T_{A} } =\frac{9900}{4900}$

$\frac{T_{B} }{T_{A} }=$ 2.02 = 2 (Approx.)

Now -

Assume that the energy flux of Star A is E(A) and that of Star B is E(B). Then -

$$\frac{E(B)}{E(A)} = \frac{\sigma\times T(B)^{4} }{\sigma \times T(A)^{2} }$

E(B) = E(A) x $(\frac{T(B)}{T(A)} )^{4}$

E(B) = E(A) x $2^{4}$

E(B) = 16 E(A)

Hence, the Energy flux from Star B is 16 times of the energy flux from Star A.

To learn more about Stars, visit the link below-

brainly.com/question/13451162

#SPJ4

4 0

2 years ago

Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2

Nookie1986 [14]

<h2>Answers:</h2>

According to Newton's Law of Gravitation, the Gravity Force is:

$F=\frac{GMm}{{r}^{2}}$ (1)

This expression can also be written as:

$F=GMm{r}^{-2}$ (2)

If we derive this force $F$ respect to the distance $r$ between the two masses:

$\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr$ (3)

Taking into account $GMm$ are constants:

$\frac{dF}{dr}dFdr=-2GMm{r}^{-3}$ (4)

Or

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

<h2> (b) dF/dr represents the rate of change of the force with respect to the distance between the bodies. </h2><h2 />

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

As the distance increases, the Force decreases.

<h2>(c) The minus sign indicates that the bodies are being forced in the negative direction. </h2>

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit $N/km$ :

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

We have a force that decreases with a rate 1 $\frac{dF_{1}}{dr}dFdr=4N/km$ when $r=20000km$ :

$4N/km=-2\frac{GMm}{{(20000km)}^{3}}$ (6)

Isolating $-2GMm$ :

$-2GMm=(4N/km)({(20000km)}^{3})$ (7)

In addition, we have another force that decreases with a rate 2 $\frac{dF_{2}}{dr}dFdr=X$ when $r=10000km$ :

$XN/km=-2\frac{GMm}{{(10000km)}^{3}}$ (8)

Isolating $-2GMm$ :

$-2GMm=X({(10000km)}^{3})$ (9)

Making (7)=(9):

$(4N/km)({(20000km)}^{3})=X({(10000km)}^{3}$ (10)

Then isolating $X$ :

$X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}$

Solving and taking into account the units, we finally have:

$X=-32N/km$ >>>>This is how fast this force changes when $r=10000 km$

7 0

4 years ago

Read 2 more answers

A car accelerates uniformly from 5m/s to 15m/s taking 7.5 seconds. How far did it travel during this period

vodomira [7]

The required answer is 66.925 m which is the distance travelled by the car.

4 0

3 years ago

Read 2 more answers

Calculate the heat flux (in W/m2) through a sheet of a metal 11-mm thick if the temperatures at the two faces are 350 and 110 ˚C

Masja [62]

q = 1156363.6W/m².

To calculate the heat flux per unit area (W/m²) of a sheet made of metal:

q = -k(ΔT/Δx)

q = -k[(T₂ - T₁)/Δx]

Where, k is the thermal conductivity of the metal, ΔT is the temperature difference and Δx is the thick.

With Δx = 11 mm = 11x10⁻³m, T₂ = 350°C and T₁ = 110°C, and k = 53.0 W/m-K:

q = -53.0W/m-K[(110°C - 350°C)/11x10⁻³m

q = 1156363.6W/m²

3 0

3 years ago

Oxygen at 50 lbf/in.2, 200 F is in a piston/cylinder arrangement with a volume of 4 ft3. It is now compressed in a polytropic pr

Ksju [112]

Answer:

The value of heat transfer during the process Q = - 29.49 KJ

Explanation:

Given data

$P_{1}$ = 50 $\frac{lbf}{in^{2} }$ = 344.7 k pa

$V_{1} = 4 ft^{3}$ = 0.113 $m^{3}$

$T_{1} = 200$ F = 366.4 K

$T_{2} = 400 F$ = 477.6 K

Poly tropic index n = 1.2

gas constant for oxygen = 0.26 $\frac{KJ}{kg K}$

From ideal gas equation

$P_{1}$ $V_{1}$ = m R $T_{1}$

Put all the values in above equation we get

⇒ 344.7 × 0.113 = m × 0.26 × 366.4

⇒ m = 0.408 kg

Heat transfer in poly tropic process is given by

Q = $\frac{\gamma - n}{( \gamma - 1)( n - 1)} [ {m R (T_{1} - T_{2} ) ]$

Put all the values in above formula we get

⇒ Q = $\frac{1.4 - 1.2}{( 1.4 - 1)( 1.2 - 1)} [ {m R (T_{1} - T_{2} ) ]$

⇒ Q = 2.5 × 0.408 × 0.26 × ( 366.4 - 477.6 )

⇒ Q = - 29.49 KJ

This is the value of heat transfer during the process & negative sign shows that heat is lost during the process.

3 0

4 years ago