Answer:
The portfolio should invest 48.94% in equity while 51.05% in the T-bills.
Explanation:
As the complete question is not given here ,the table of data is missing which is as attached herewith.
From the maximized equation of the utility function it is evident that

For the equity, here as
is percentage of the equity which is to be calculated
is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
is the risk aversion factor which is given as 4.
is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59
By substituting values.

So the weight of equity is 48.94%.
Now the weight of T bills is given as

So the weight of T-bills is 51.05%.
The portfolio should invest 48.94% in equity while 51.05% in the T-bills.
Answer:
(a) 61.25 N
(b) 6.25 kg
(c) 6.25 Kg
Explanation:
Weight on moon = 10 N
Acceleration due to gravity on moon = 1.6 m/s^2
Acceleration due to gravity on earth = 9.8 m/s^2
Let m be the mass of the package.
(a) Weight on earth = mass x acceleration due to gravity on earth
Weight on earth = 6.25 x 9.8 = 61.25 N
(b) Weight on moon = mass x acceleration due to gravity on moon
10 = m x 1.6
m = 6.25 kg
(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.
A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
Answer:

Explanation:
The density changes means that the length in the direction of the motion is changed.
Therefore,

Given :
Side, b = h = 0.13 m
Mass, m = 3.3 kg
Density = 8100 
So,


l = 0.024 m
Then for relativistic length contraction,







Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).