Explanation:
Equilibrium position in y direction:
W = Fb (Weight of the block is equal to buoyant force)
m*g = V*p*g
V under water = A*h
hence,
m = A*h*p
Using Newton 2nd Law

Hence, T time period
T = 2*pi*sqrt ( h / g )
Thanks for sharing that information. After extensive calculation,
we can say with assurance that after some number of seconds,
a loud "crunch" is perceived by the souls aboard the ill-fated vessel.
Answer:
The farther star will appear 4 times fainter than the star that is near to the observer.
Explanation:
Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time
Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)
This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by
Hence we sense it as 4 times fainter than the nearer star.
<u>Answer:</u>
The height of ramp = 124.694 m
<u>Explanation:</u>
Using second equation of motion,

From the question,
u = 31 m/s; s = 156.3 m, a=0
substituting values

t = 
= 5.042 s
Similary, for the case of landing
t = 5.042 s; initial velocity, u =0
acceleration = acceleration due to gravity, g = 9.81 
Substituting in 

h = 124.694 m
So height of ramp = 124.694 m