Suppose you were marooned on a tropical island and had to use seawater (density 1.10 g·cm3) to make a primitive barometer. what height would the water reach in your barometer when a mercury barometer would reach 73.5 cm? the density of mercury is 13.6 g·cm3

We know that P = hρg

Where:

P - pressure Pa,

h - height in meter,

ρ – would be the density in kg / m^3; and

g - acceleration due to gravity is m / s^2

p = hρg if h = 0.735 meter, ρ = 13600 kg / meter^3, g = 10 meter/ sec^2

p = 0.735*13600*10 = 99960 Pa or 

P = 1 x 10^5 Pa

Now with sea water if we have to make a barometer:

ρ = 1100 kg / meter^3 (given) 

P = hρg if we put the value of P calculated and the value of ρ = 1100 kg / meter^3 given, we will 

get, 1 x 10^5 = h*1100*10

therefore, h = 9.09 meter or 29.82 feet of water.

an obstruction

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.