An air bubble released by a remotely operated underwater vehicle, 120 m below the surface of a lake, has a volume of 1.40 cm3. the surface of the lake is at sea level, and the density of the lake water can be approximated as that of pure water. as the bubble rises to the surface, the temperature of the water and the number of air molecules in the bubble can each be approximated as constant. find the volume (in cm3) of the bubble just before it pops at the surface of the lake.

17.7 cm^3

Explanation:

depth, h = 120 m

density of water, d = 1000 kg/m^3

V1 = 1.4 cm^3

P1 = P0 + h x d x g

P2 = P0

where, P0 be the atmospheric pressure

Let V2 be the volume of the bubble at the surface of water.

P0 = 1.01 x 10^5 Pa

P1 = 1.01 x 10^5 + 120 x 1000 x 9.8 = 12.77 x 10^5 Pa

Use

P1 x V1 = P2 x V2

12.77 x 10^5 x 1.4 = 1.01 x 10^5 x V2

V2 = 17.7 cm^3

Thus, the volume of bubble at the surface of water is 17.7 cm^3.

heats surface water up

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.