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GrogVix [38]
3 years ago
11

We would be more likely to get an accurate measurement of the distance from Earth to a nearby star if we took the angular measur

ements at times separated by what time interval?

Physics
1 answer:
NeTakaya3 years ago
8 0

Answer:

6 month interval

Explanation:

The distance to a nearby star in theory is more simple than

one might think! First we must learn about the parallax effect. This is the mechanism our eyes use to perceive things at a distance! When we look at the star from the earth we see it at different angles throughout the earth's movement around the sun similar to how we see when we cover on eye at a time. Modern telescopes and technology can help calculate the angle of the star to the earth with just two measurements (attached photo!) Since we know the distance of the earth from the sun we can use a simple trigonometric function to calculate the distance to the star. The two measurements needed to calculate the angle of the star to the earth caused by parallax (in short angle θ) are shown in the second attached photo.

So using a simple trigonometric function Sin\theta=\frac{r}{d} we can solve for d which is the distance of the earth to the star:

d=\frac{r}{Sin\theta}

In the first attached photo a picture where r is the distance to the star and the base of the triangle is the diameter of the earth.

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7 0
2 years ago
In an old house, the heating system uses radiators, which are hollow metal devices through which hot water or steam circulates.
RSB [31]

Answer:

T = 125.30C

Explanation:

Using Stefan-Boltzmann law

P/A = e* Stefan-Boltzmann constant * T^4

Where p= radiant power in watts

e = emissivity

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A= Surface area of the object measured in m^2

Considering that after painting the object have the radiant power then

0.83* A* Stefan-Boltzmann constant *(337.1^4) = 0.46 * A* Stefan-Boltzmann constant * Tnew^4

Cancelling the common terms yields

0.83*(337.1^4) = 0.426*Tnew^4

Fourthroot(0.83*(337.1^4)) = Tnew

Tnew = 398.3k = in degree Celsius 125.3

8 0
3 years ago
Please help. Not my strongest subject
Mars2501 [29]
A) You'll notice from "c)" that the distance the object travelled was 75m, so you get V_a_v_e_r_a_g_e = 75m/15s = 5 m/s.

b) a = dV/dT, so a = (0-10)/15 and then you get a = -2/3 m/s ^{2}.

c) S = S_0 + V_0T + 0,5aT^2 and you get S = 10T - (2/6)T^2 and you plug in T = 15s and you'll get S = 75m
4 0
3 years ago
Which of the following locations will be relatively warmer during summers, and why? A. An inland location, because water heats u
Sonbull [250]
D - An inland location, because land has lower specific heat than water

Explanation:

The specific heat is a measure of how much heat must be supplied to warm up a substance by a certain number of degrees.
Water has a much higher specific heat than land, so it takes much more heat energy to heat up a given mass of water compared to the land.
To say it another way, the specific heat of land is much less than that of water, so it doesn't take as much heat energy to warm up the land.
Thus, an inland location will be relatively warmer during summers.
3 0
3 years ago
A wave with a period of 0.008 second has a frequency of
rodikova [14]
D. 125 Hz
f =   \frac{1}{t}  =  \frac{1}{0.008}  \\  = 125hz
4 0
3 years ago
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