Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.
where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.
Let the orbital period of the earth be 
and its mean distance of from the sun be 
.
Also let the orbital period of the planet be 
and its mean distance from the sun be 
.
Equation (2) therefore implies the following;
We make the period of the planet the subject of formula as follows;
But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore
Substituting equation (5) into (4), we obtain the following;
cancels out and we are left with the following;
Recall that the orbital period of the earth is about 365.25 days, hence;
The answer is 60 mph.
The speed (v) is distance (d) per time (t): v = d/t
Car A:
v1 = ?
t1 = 2 h
d1 = ?
___
v1 = d1/t1
d1 = v1 * t1
Car B:
v2 = ?
t2 = 1.5 h
d2 = ?
___
v2 = d2/t2
d2 = v2 * t2
<span>Two cars traveled equal distances:
d1 = d2
</span>v1 * t1 = v2 * t2
<span>Car B traveled 15 mph faster than Car A:
v2 = v1 + 15
</span>v1 * t1 = v2 * t2
v2 = v1 + 15
________
v1 * 2 = (v1 + 15) * 1.5
2v1 = 1.5v1 + 22.5
2v1 - 1.5v1 = 22.5
0.5v1 = 22.5
v1 = 22.5/0.5
v1 = 45 mph
v2 = v1 + 15
v2 = 45 + 15
v2 = 60 mph
As the water russhes toward the shore, it rises because it is pushing against it.<span />
A testing instrument that's used to measure electrical signals
in a circuit and display them as waveforms on a screen is called
an oscilloscope.
