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Igoryamba
3 years ago
12

20. Consider two Stars A and B with temperatures Ta and TB and radii Ra and RB respectively. If TA

Physics
1 answer:
Rus_ich [418]3 years ago
4 0

Answer:

L_A = 9 L_B

Explanation:

The formula that relates the luminosity of a star (L) to its radius (R) and the temperature (T) is

L=\frac{R^2}{T^4}

For star B, we can write:

L_B=\frac{R_B^2}{T_B^4}

For star A, we have

T_A = 2 T_B\\R_A = 12 R_B

So the luminosity of star A is

L_A=\frac{R_A^2}{T_A^4}=\frac{(12 R_B)^2}{(2 T_B)^4}=\frac{144 R_B^2}{16 T_B^4}=9\frac{R_B^2}{T_B^4}=9 L_B

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ns:Select the correct answer. In a video game, a flying coconut moves at a constant velocity of 20 meters/second. The coconut hi
Len [333]
You would need to use the equation a= (v-u)÷t
You need to substitute in the correct numbers.
a= (10-20)÷1
Your answer is -10m/s^2
5 0
3 years ago
Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

8 0
3 years ago
Can some one help me with this question its hard :( down below ill give brainliest
tensa zangetsu [6.8K]
I’m pretty sure it’s c sorry if I’m wrong
3 0
3 years ago
The three types of nuclear radiation<br>​
Eva8 [605]
The three types are alpha beta and gamma
6 0
3 years ago
the train accelerates from 30 km/h to 45 km/h in 15 secs. a. find its acceleration. b. distance it travels during this time ...
Sonbull [250]
Acceleration = v-u/t
= (45-30)/15
= 1 km/h


Distance = ut + 1/2 at^2
s = [30 x 15]  + 1/2 x 1 x 15^2 
= 562.5 km

Hope this helps

3 0
3 years ago
Read 2 more answers
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