Answer:
Explanation:
Given:
temperature of source reservoir, 
temperature of sink reservoir, 
energy absorbed from the source, 
work done, 
a.
<u>Now change in entropy of the surrounding:</u>
<em>Since heat engine is a device that absorbs heat from a high temperature reservoir and does some work giving out heat in the universe as the byproduct.</em>
b.
<u>We know Carnot efficiency is given as:</u>
<u>Now the Carnot work done:</u>
.......................(1)
c.
From eq. (1) we have the Carnot work, so the difference:
Now, we find:
Let's be clear: The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.
Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.
After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.
I = MR^2
The Attempt at a Solution:::
I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2
I total = 3ML^2/2
It says the answer is 3ML^2/4 though.
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Answer:
Every dielectric medium will have a dielectric (strength) constant.
For space / vacuum it is 1. For rest of the dielectric media, it is greater than 1. The capacitance of a capacitor filled with a material of dielectric constant k, is given by the relation:
C = k . C0 (where C0 is the capacitance with space / vacuum as the dielectric material in that capacitor which means no dielectric between the plates)
So, it clearly states that the higher the dielectric constant, the more the capacitance i.e. it is directly proportional to the dielectric strength.
Dielectric constant varies with temperature.
Explanation:
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Answer:
Star A would be brighter than Star B
Explanation:
The apparent brightness of a star as perceived on Earth is dependent on its temperature, size, luminosity and distance from the Earth. Apparent brightness is the visible brightness to the eye at the surface of the Earth, while luminosity is the true brightness at the surface of the star.
A hotter star will radiate more energy per second per meter square of surface area. A larger star will have a greater surface area for radiation of energy, thus increasing the luminosity. For two identical stars, the difference in apparent brightness will be dependent on their distances from Earth.
Brightness and distance from earth have an inverse square relationship.
∝
Assuming the star is a point source of radiation, as distance from the source is increased, the radiation is distributed over a surface proportional to the distance form the source. As distance is further increased, the radiation is distributed over a larger surface area reducing the effective luminosity.
If one star (Star B) is twice as far from the earth as the first (Star A), the brightness of Star B will be 
of Star A.
Thus, Star B will appear to be a quarter of the brightness of Star A. Or, Star A will appear to be 4 times as bright as Star B.
