Answer:
E = 1/2 M V^2 + 1/2 I ω^2 = 1/2 M V^2 + 1/2 I V^2 / R^2
E = 1/2 M V^2 (1 + I / (M R^2))
For a cylinder I = M R^2
For a sphere I = 2/3 M R^2
E(cylinder) = 1 + 1 = 2 omitting the 1/2 M V^2
E(sphere) = 1 + 2/3 = 1.67
E(cylinder) / E(sphere) = 2 / 1.67 = 1.2
The cylinder initially has 1.20 the energy of the sphere
The PE attained is proportional to the initial KE
H(sphere) = 2.87 / 1/2 = 2.40 m since it has less initial KE
Answer: 
Explanation:
Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period 
of a planet is proportional to the cube of the semi-major axis 
of its orbit”:
(1)
Now, if is measured in years (Earth years), and is measured in astronomical units (equivalent to the distance between the Sun and the Earth: 
), equation (1) becomes:
(2)
So, knowing 
and isolating from (2) we have:
![a=\sqrt[3]{T^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7BT%5E%7B2%7D%7D)
(3)
![a=\sqrt[3]{(619.36 years)^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%28619.36%20years%29%5E%7B2%7D%7D)
(4)
Finally:
T
his is the distance between the dwarf planet and the Sun in astronomical units
Converting this to kilometers, we have:
The answers to question 4 d
Answer:
b.only when the current in the first coil changes.
Explanation:
An induced current flow in the second coil only when there is a change in current in the first cool. A steady current will produce no change in flux (due to magnetic effect of a current) by the first coil, and according to Faraday, induced current is only produced when there is a change in flux linkage.
