Answer: A
Explanation:
The acceleration of gravity is always 9.8 m/s^2 downwards, even if the velocity is 0 m/s.
Answer:
The rate at which bus 1 is going is 55 mph
The rate at which bus 1 is going is 35 mph
Explanation:
As per the question:
Suppose, the distance traveled by Bus 1 be 'd' at the rate R after a time, t = 3h
Thus
Suppose, the distance traveled by Bus 1 be 'd'' at the rate, R'20 mph slower than the rate of Bus 1 after the same time.
R' = R - 20
The distance is given as the product of rate and time:
d = Rt (1)
Now, the total distance given is 270 miles:
d + d' = 270
Now, using eqn (1):
Rt + R't = 270
3(R + R - 20) = 270
6R = 270 + 60
R = 55 mph
R' = R - 20 = 55 - 20 = 35 mph
Answer:
Explanation:
The period of the comet is the time it takes to do a complete orbit:
T=1951-(-563)=2514 years
writen in seconds:
Since the eccentricity is greater than 0 but lower than 1 you can know that the trajectory is an ellipse.
Therefore, if the mass of the sun is aprox. 1.99e30 kg, and you assume it to be much larger than the mass of the comet, you can use Kepler's law of periods to calculate the semimajor axis:
![T^2=\frac{4\pi^2}{Gm_{sun}}a^3\\ a=\sqrt[3]{\frac{Gm_{sun}T^2}{4\pi^2} } \\a=1.50*10^{6}m](https://tex.z-dn.net/?f=T%5E2%3D%5Cfrac%7B4%5Cpi%5E2%7D%7BGm_%7Bsun%7D%7Da%5E3%5C%5C%20a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGm_%7Bsun%7DT%5E2%7D%7B4%5Cpi%5E2%7D%20%7D%20%5C%5Ca%3D1.50%2A10%5E%7B6%7Dm)
Then, using the law of orbits, you can calculate the greatest distance from the sun, which is called aphelion:
