Well, the density of the water is
so i believe that is what the question is asking for :)
Actually, they're not. There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around. And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night.
Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.
Why does the night sky change at various times of the year ? Here's how to
think about it:
The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.
Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.
In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ? Now they're straight in the
direction of the sun. So they're directly overhead at Noon, not at Midnight.
THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.
Answer:
Explanation:
The rotation rate of the man is:
The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:
![(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega](https://tex.z-dn.net/?f=%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%5Ccdot%20%280.16%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%29%20%3D%20%5B%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%2B20000%5C%2Ckg%5Ccdot%20m%5E%7B2%7D%5D%5Ccdot%20%5Comega)
The final angular speed is:
Answer:
Tangential acceleration is in the direction of velocity - along the circumference of a circle if the object is undergoing circular motion
a = (V2 - V1) / T
Radial acceleration is perpendicular to the direction of motion if the object is not moving in a straight line (perhaps along the circumference of a circle)
a = m V^2 / R = m ω^2 R where R is the radius vector of the velocity - note that the Radius vector is directed from the center of motion to the object and for circular motion would be constant in magnitude but not in direction
The answer should be A) solar power
