An air bubble originating from an under water diver has a radius of 5 mm at some depth h. when the bubble reaches the surface of the water, it has a radius of 6.6 mm. assuming the temperature of the air in the bubble remains constant, determine the absolute pressure at this depth h. the acceleration of gravity is 9.8 m/s 2 . answer in units of pa.

Absolute Pressure at depth h =

Explanation:

Given:

Assumptions:

According to Ideal Gas Law,

where,

For the given case, the temperature of the air bubble is constant. The number of molecules of the bubble also does not change, therefore,

PV = constant.

The pressure at the surface is equal to the atmospheric pressure,

The pressure at depth h is equal to sum of atmospheric pressure and the pressure of the water upto depth h,

Using these values,

The absolute temperature at that depth is given by

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft