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.

b

explanation:

bc its just b