Light rays from stars bend toward smaller angles as they enter Earth's atmosphere. a. Explain why this happens using Snell's law
and the speed of light. b. Where are the actual stars in relation to their apparent position as viewed from the Earth's surface?
1 answer:
Answer:
Following are the answer to this question:
Explanation:
In option (a):
- The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.
- Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.
In option (b):
- Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.
- Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.
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Answer:
ºC
Explanation:
First, let's write the energy balance over the duct:
It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:
The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.
So, let's isolate :
The Cp of the air at 27ºC is 1007 (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are
and Q.
Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.
The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:
Perimeter:
Surface area:
Then, the heat Q is:
Finally, find the exit temperature:
=27.0000077 ºC
The temperature change so little because:
- The mass flow is so big compared to the heat flux.
- The transfer area is so little, a bigger length would be required.
Answer:
b) TA = TB = TC
Explanation:
- When put in contact each other, and isolated, both blocks will exchange heat till they reach to thermal equilibrium.
- During this process, the body at a higher temperature, will loss heat, tat it will be gained by the other body.
- The equilibrium condition will be reached when the following equation be met:
- Replacing by the values of T₀A = 300º C, and T₀B = 400ºC, and simplifying common terms as mA = mB, we can solve for Tfin, as follows:
- So, when both blocks reach to equilibrium, they will be at a common final temperature, 350ºC.
- When put in contact with block C, at the same temperature, at that instant, the three blocks will have the same common temperature of 350 ºC.
- So, option b) is the right one.
Answer:
- The total distance traveled is 28 inches.
- The displacement is 2 inches to the east.
Explanation:
Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector pointing in the west direction, the ant start at position
Then, moves to
so, the distance traveled here is
after this, the ant travels to
so, the distance traveled here is
The total distance traveled will be:
The displacement is the final position vector minus the initial position vector:
This is 2 inches to the east.