Answer:
21.35 cm^3
Explanation:
let the volume at the surface of fresh water is V.
The volume at a depth of 100 m is V' = 2 cm^3
temperature remains constant.
density of water, d = 1000 kg/m^3
Pressure at the surface of fresh water is atmospheric pressure,
P = Po = 1.013 x 10^5 N/m^2
The pressure at depth 100 m is P' = Po + hdg
P' =
P' = 10.813 x 10^5 N/m^2
Use the Boyle's law
P V = P' V'
V = 21.35 cm^3
Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.
Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
And the distance traveled downwards is:
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
Replacing into the first equation:
Rationalizing:
Let's call v1 the final speed of the package dropped from a height H. Thus:
Let v2 be the final speed of the package dropped from a height 4H. Thus:
Taking out the square root of 4:
Dividing v2/v1 we can compare the final speeds:
Simplifying:
The final speed of the second package is twice as much as the final speed of the first package.