Answer:
Explanation:
8. Which process(es) of the water cycle—precipitation, evaporation, condensation, runoff, percolation or transpiration-might contribute to the addition of pollutants to rivers, lakes, and oceans? Why? Precipitation and runoff would be the most responsible processes.
An uncharged body is a body that has equal magnitude of positive as well as negative charge on it. So as soon as some charged body is placed near it, the charged body induces negative charges on the uncharged body.
In <u>370 K to 375 K </u>temperature intervals of 5 K, would be the greatest increase in the entropy of the sample.
Option: C
<u>Explanation</u>:
Because the largest difference in molar entropy occurs when a condensed phase (solid/liquid) transforms to the gas phase. Then change in entropy is equal to heat transfer divided by temperature: 
.
According to given ice sample at 260 K, when this solid sample start converting into liquid sample it will gain positive temperature and steam will take place near 373 K (273 K ice temperature + 
temperature of boiling water). Therefore it’s very obvious that greatest increase in entropy will occur during 370 K – 375 K.
When standing behind the sound source
<span>373.2 km
The formula for velocity at any point within an orbit is
v = sqrt(mu(2/r - 1/a))
where
v = velocity
mu = standard gravitational parameter (GM)
r = radius satellite currently at
a = semi-major axis
Since the orbit is assumed to be circular, the equation is simplified to
v = sqrt(mu/r)
The value of mu for earth is
3.986004419 Ă— 10^14 m^3/s^2
Now we need to figure out how many seconds one orbit of the space station takes. So
86400 / 15.65 = 5520.767 seconds
And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity
2 pi r / 5520.767
Finally, combining all that gets us the following equality
v = 2 pi r / 5520.767
v = sqrt(mu/r)
mu = 3.986004419 Ă— 10^14 m^3/s^2
2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r)
Square both sides
1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r
Multiply both sides by r
1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2
Divide both sides by 1.29527 * 10^-6 s^2
r^3 = 3.0773498781296 * 10^20 m^3
Take the cube root of both sides
r = 6751375.945 m
Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So
6751375.945 m - 6378137.0 m = 373238.945 m
Converting to kilometers and rounding to 4 significant figures gives
373.2 km</span>
