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Vedmedyk [2.9K]
2 years ago
8

Suppose a star has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 ) but is located at a distance of 300 ligh

t-years. What is its luminosity?
Physics
1 answer:
iragen [17]2 years ago
6 0

For  a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as

L=2.7*10^30w

<h3> What is its luminosity?</h3>

Generally, the equation for the luminosity  is mathematically given as

L=4*\pi^2*b

Therefore

L=4*\pi^2*b

L=4* \pi *(2.83*10^{18})*2.7*10^{-8}

L=2.7*10^30

In conclusion, the luminosity

L=2.7*10^30w

Read more about Light

brainly.com/question/25770676

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 net force is mass multiplied by acceleration. hope this helps
3 0
3 years ago
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Need some help
Keith_Richards [23]

The answer is C) the density of the rock

Density of rock is the dependent variable, because it depends on the temperature. The temperature can’t be the dependent variable because ,the density of a rock does not have magical powers that can change temperature of a room. However changing the temperature of the room ,will change the density of the rock. Hope this helps !

4 0
3 years ago
A person driving her car at 43 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow
siniylev [52]

Answer:

If she hits the brakes, she will travel 13m before stopping. If she hits the gas, she will travel 28 before the light turns red. She should try to stop

Explanation:

To know how far she will travel before stopping, we need to use a kinematic formula which initial final velocity (Vf), initial velocity (V0), acceleration (a) and distance traveled(x):

V_{f}^{2}=V_{0} ^{2} +2ax

The moment in which she stops is when the final velocity equals zero. In this case, initial velocity is 43km/h=12m/s, and its maximum deceleration is -5.4m/s^2. Plugging in these values and solving for x:

0^2=(12m/s)^2+2(-5.4m/s)x\\x=13m

She will travel 13m before stopping

If she hits the gas, we need another kinematic formula, which relates distance traveled, initial speed, time (t) and acceleration.

x=v_{0}t+0.5a*t^2

If we know her car can accelerate from 43km/h=12m/s to 70km/h=19m/s in 8.1 s, we can know its acceleration:

a=\frac{19m/s-12m/s}{8.1s}=0.86m/s^2

In this case, the time before the light turns red is 2.0s. Plugging in all those values:

x=12m/s*2s+0.5*0.86m/s^2*(2s)^2=28m

If she hits the gas, she will travel 28m before the light turns red. She won't even reach the intersection, so she would try to stop.

5 0
3 years ago
A force p is 46.67N. if it inclined at an angle of 50°, find its horizontal component​
Ivanshal [37]

Explanation:

F_{x} = 46.67 N(cos 50°) = 30.0 N

8 0
3 years ago
Unpolarized light of intensity 0.0288 W/m2 is incident on a single polarizing sheet. What is the rms value of the electric field
galina1969 [7]

Answer:

the rms value of the electric field component transmitted is 3.295 V/m

Explanation:

Given;

intensity of the unpolarized light, I = 0.0288 W/m²

For unpolarized light, the relationship between the amplitude electric field and intensity is given as;

E_{max} = \sqrt{2\mu_0cI} \\\\E_{max} = \sqrt{2(4\pi \times 10^{-7})(3\times 10^8)(0.0288)} \\\\E_{max} = 4.66 \ V/m

The relationship between the rms value of the electric field and the amplitude electric field is given as;

E_{rms} = \frac{E_0}{\sqrt{2} } =\frac{E_{max}}{\sqrt{2} }  \\\\E_{rms} = \frac{4.66}{\sqrt{2} }\\\\E_{rms} = 3.295 \ V/m

Therefore, the rms value of the electric field component transmitted is 3.295 V/m

5 0
3 years ago
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