Answer:
i can't click the answer bottom but the answer is "17th to 18th century" i hope this helps
Answer:
hope this helps!
Explanation:
Volume of the air bubble, V1=1.0cm3=1.0×10−6m3
Bubble rises to height, d=40m
Temperature at a depth of 40 m, T1=12oC=285K
Temperature at the surface of the lake, T2=35oC=308K
The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa
The pressure at the depth of 40 m: P1=1atm+dρg
Where,
ρ is the density of water =103kg/m3
g is the acceleration due to gravity =9.8m/s2
∴P1=1.103×105+40×103×9.8=493300Pa
We have T1P1V1=T2P2V2
Where, V2 is the volume of the air bubble when it reaches the surface.
V2=
Answer:
6370 J
Explanation:
By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m

where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference
