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Gelneren [198K]
3 years ago
10

A measure of the amount of light received on Earth is a star's ____.

Physics
2 answers:
Nina [5.8K]3 years ago
7 0

Answer:

a. apparent magnitude

Explanation:

Apparent magnitude is a measure of the brightness of a star (that is, amount of light) received on Earth. Star's apparent magnitude depends on:

  • star luminosity
  • star distance from Earth
  • extinction of the light, for example, caused by interstellar dust.

Helen [10]3 years ago
4 0

Apparent magnitude because Apparent is kinda like when something is obvious. So that's how I think of it.

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Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper
vodka [1.7K]

Explanation:

Expression to calculate thermal resistance for iron (R_{I}) is as follows.

             R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}  

where,   L_{I} = length of the iron bar

             k_{I} = thermal conductivity of iron

             A_{I} = Area of cross-section for the iron bar

Thermal resistance for copper (R_{c}) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where,  L_{c} = length of copper bar

             k_{c} = thermal conductivity of copper

            A_{c} = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

           Q = \frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

 Putting the given values into the above formula as follows.

       Q = \frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

  = \frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}

           = 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

                 P = \frac{Q}{T}

Here, T is 1 second so, power conducted is equal to heat transferred.

So,           P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

7 0
3 years ago
Which of these will be the correct relationship between work input and work output?
dsp73

<u>Answer:</u>

Work input = Work output * Work against friction is your answer so C

<u>Explanation:</u>

I hope this helps you :)

8 0
2 years ago
What is the driving force of charge around a circuit
Virty [35]
A force of charge that drive around a circuit is call electeons
5 0
3 years ago
The average speed between earth and the sun is 1.50 x10^8 km. Calculate the average speed of the Earth in its orbit in kilometer
cluponka [151]

Answer:

The average speed of the earth in its orbit is 29.86km/s

Explanation:

The average distance between the Earth and the Sun is 1.50x10^{8} km.

The average speed of the earth in its orbit can be found by the next equation :

v = \frac{2 \pi r}{T}  (1)

Where r is the radius and T is the period.

In this case, the orbit of the Earth can be considered as a circle

(r = 1.50x10^{8}km) instead of an ellipse.

It takes 1 year to the Earth to make one revolution around the Sun. Therefore, its period will be 365.25 days.

Notice that to express the period in terms of seconds, the following is needed:

365.25d . \frac{86400s}{1d} ⇒ 31557600s

Then, equation 1 can be used:

v = \frac{2 \pi (1.50x10^{8}km)}{31557600s}

v = 29.86km/s

7 0
3 years ago
Read 2 more answers
The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length
Zarrin [17]

The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

y_r=1.22\frac{\lambda f}{d}

Where,

y_r = Range of the radius

\lambda = wavelength

f= focal length

Our values are given by,

State 1:

d=2.00mm = 2*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 750nm = 750*10^{-9}m

State 2:

d=8.00mm = 8*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 390nm = 390*10^{-9}

Replacing in the first equation we have:

y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}

y_{r1}= 11.4\mu m

And also for,

y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}

y_{r2} = 1.49\mu m

Therefor, the airy disk radius ranges from 1.49\mu m to 11.4\mu m

7 0
3 years ago
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