Ok, assuming "mj" in the question is Megajoules MJ) you need a total amount of rotational kinetic energy in the fly wheel at the beginning of the trip that equals
(2.4e6 J/km)x(300 km)=7.2e8 J
The expression for rotational kinetic energy is
E = (1/2)Iω²
where I is the moment of inertia of the fly wheel and ω is the angular velocity.
So this comes down to finding the value of I that gives the required energy. We know the mass is 101kg. The formula for a solid cylinder's moment of inertia is
I = (1/2)mR²
We want (1/2)Iω² = 7.2e8 J and we know ω is limited to 470 revs/sec. However, ω must be in radians per second so multiply it by 2π to get
ω = 2953.1 rad/s
Now let's use this to solve the energy equation, E = (1/2)Iω², for I:
I = 2(7.2e8 J)/(2953.1 rad/s)² = 165.12 kg·m²
Now find the radius R,
165.12 kg·m² = (1/2)(101)R²,
√(2·165/101) = 1.807m
R = 1.807m
Answer:
B. About 12 degrees
Explanation:
The orbital period is calculated using the following expression:
T = 2π*(
)
Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.
Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.
Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:
= 
Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be 
years, or 31.62 years.
We find the amount of degrees it moves in 1 year:
or about 12 degrees.
It lasts 29 1/2 days.
-the sidereal month- is the true period of the moon's revolution around Earth. It lasts 27 1/3 days.
-the difference of 2 days between the synodic and sidereal cycles is due to the Earth- moon system also moving in an orbit around the sun.
One that can help you is:
ΔT=<span>T<span>Final</span></span>−<span>T<span>Initia<span>l
That is of course adding both tmepratures. There is one more that is a lil bit more complex
</span></span></span><span><span>Tf</span>=<span>Ti</span>−Δ<span>H<span>rxn</span></span>∗<span>n<span>rxn</span></span>/(<span>C<span>p,water</span></span>∗<span>m<span>water</span></span>)
This one is taking into account that yu can find temperature and that there could be a change with a chemical reaction. Hope this helps</span>
