radius = .4 m so piston area = pi(.16) =.506 m^2
(a) is a change in volume problem all apparently at constant temperature = 293 K
First new volume with the piston:
P1V1/T1 = P2V2/T2
T1=T2 = 293 K
V2 = V1 (P1/P2)
P1 = 1 atm = 10^5 Pascals
P2 = 1 atm + (21*9.8)/.506 = 10^5+407
V2/V1 = 100000/100407 =.996
H2 = .996 (50) = 49.8 cm hardly moves
now add the man
V3 = V1(P1/P3)
P3 = 1 atm + (100*9.8)/.506 =10^5+1937
V3/V1 = 100000/101937 = .981
H3 = 50*.981 = 49.05 cm
when the piston was added the level went down to 49.8 and then it went down to 49.05 when the man stepped on so it went down about .8 cm due to the man.
For Part V
V2/V1 = 50/49.05 = 1.02
constant pressure so temp goes up
P1V1/T1 = P2 V2/T2
V2/V1 = T2/T1
T1 = 293 K
1.02 = T2/293
T2 = 298.7 K = 25.7 C
thank you so much! it's really clear now with the way you've explained it
two airplanes leave an airport at the same time. the velocity of the first airplane is 700 m/h at a heading of 52.2degrees . the velocity of the second is 600 m/h at a heading of 97 degrees . how far apart are they after 3.1 h? answer in units of m.
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