An m = 7.25 kg mass is suspended on a string which is pulled upward by a force of F = 76.7 N as shown in the figure. If the upwa
rd velocity of the mass is 2.25 m/s right now, then what is the velocity 3.50 s later?
1 answer:
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- initial velocity=u=0m/s
- Final velocity=v=25m/s
- Time=t=30s
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.