An air bubble is expelled by an underwater diver at a depth of 22.3 m. as the bubble rises to the surface, the pressure decreases and the volume of the bubble therefore increases. the temperature of the water is constant. find the final volume of the bubble if its initial volume is 10.5 cm3.

s

V₂ = 27.33 cm³

Explanation:

given,

depth of the air bubble under water = 22.3 m

initial volume of the bubble = 10.5 cm³

final volume of the bubble = ?

Pressure under water increase

P₁ = P₂ + ρ g h

 P₂= 10⁵ Pa

g is the acceleration due to gravity, g = 9.8 m/s²

Using idea gas law

P₁ V₁ = P₂ V₂

V₂ =

V₂ =

 V₂ = 27.33 cm³

final volume of the bubble will be equal to  V₂ = 27.33 cm³

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west