Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water.
If the cylinder is 0.4 m tall and the absolute (or total) pressure at the bottom is 1.1 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 104 kg/m^3, and the ambient atmospheric pressure to be 1.013e5 Pa)

0.05m

Explanation:

Density of water = ρ(w) = 1000 kg/ m³ ;  

Density of Mercury = ρ(m) = 13628.95 kg/ m³  

Total pressure at bottom of cylinder=1.1atm

Therefore, pressure due to water and mercury =1.1-1 =atm

0.1atm=10130pa

The pressure at the bottom is given by,

ρ(w) x g[0.4 - d] + ρ(m) x g x d  = 10130

1000 x 9.8[0.4 - d] + 13628.95 x 9.8 d = 10130

3924 - 9810d + 133416d= 10130

123606d= 6206

d= 6202/123606

d= 0.05m

Depth of mercury alone =  d = 0.05m

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form