V2 = (ρ•g•h+p)• V1/ p =
= (1000•10•50 + 100000) •15/100000 = 90 cm^3
ρ = 1000 kg/m^3 is the density of water,
p is the atm. pressure
An air bubble of volume 15 cc is formed at a depth of 50 m in a lake. If the temperature of the bubble while rising remains constant then what will be the volume of the bubble at the surface?(Given: g=10 m/s2 & atm. pressure = 1.0 * 100000 Pa)
* Physics - Elena, Sunday, April 8, 2012 at 2:05pm
p1•V1 = p2•V2,
ρ•g•h• V1 = p2•V2,
V2 = ρ•g•h• V1/ p2,
ρ = 1000 kg/m^3 is the density of water,
p is the atm. pressure
* Physics - elena plz read it, Monday, April 9, 2012 at 12:04am
i am not getting the answer which is 90cc
1 answer