Calculate the (a) liquid pressure, and (b) total pressure
on a scubadiver 6 m below the surface of a lake?
Given that atmospheric pressure, Po = 1.01 105 Pa,
and density of lake water = 10kg m-3. You may take
g= 10 m s-
Answers
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Explanation:
we know,pressure=h,r,g
where h=depth of liquid,r=density g=acceleration due to gravity
a)h=6m
r=10kgm^-3
g=10ms-
p=6×10×10=600Pa
b)total pressure=Po + hrg(p)
Po=1.01105Pa(atmospheric pressure)
tp=1.01105+600=601.01105Pa
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