Answer: 117 kPa
Explanation:
For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa
For the liquid at depth 9m, the gauge pressure is equal to= P₂
Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.
So, For finding gauge pressure we have formula P= ρ * g * h
Also gravity also remains same for both liquids
So taking ratio of their respective pressures we have
= 
So 
= 
Or P₂= 39 * 3 = 117 kPa
Answer:
t = 444.125 sec
Explanation:
Given data:
V = 24 volt
I = 0.1 ampere
mass of water mw = 51 gm
cr = 4.18 J/gm degree K^-1
mass of resistor = 8 gm
cr = 3.7 J/gm degree K^-1
we know that power is given as
Power P = VI
But P =E/t
so equating both side we have
solving for t
t = 444.125 sec
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Solving for vf gives you PiVi/Pf. Now plug in 101kPa*10L/43kPa = 23.48L. Using significant figures i would round to 23.5L
