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:
i think the planet is Uranus
There doesn't seem to be any direct connection.
Answer:
Rectangular path
Solution:
As per the question:
Length, a = 4 km
Height, h = 2 km
In order to minimize the cost let us denote the side of the square bottom be 'a'
Thus the area of the bottom of the square, A = 
Let the height of the bin be 'h'
Therefore the total area, 
The cost is:
C = 2sh
Volume of the box, V =
(1)
Total cost,
(2)
From eqn (1):

Using the above value in eqn (1):


Differentiating the above eqn w.r.t 'a':

For the required solution equating the above eqn to zero:


a = 4
Also

The path in order to minimize the cost must be a rectangle.