Calculating Gauge Pressure in a Two-Story House

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The gauge pressure of the main water line entering the house on the first floor is 2.00 x 10^5 Pa. To calculate the gauge pressure at a faucet on the second floor, 7.0 m above, the formula P2 = P1 + (density x g x h) is applied. The initial calculation yields P1 = 131330 Pa, which seems correct initially. However, an error is identified regarding the height of the pipe, as it should be considered negative since pipes typically enter through the basement. This adjustment impacts the final gauge pressure calculation at the second-floor faucet.
wallace13
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The main water line enters a house on the first floor. The line has a gauge pressure of 2.00 x10^5 Pa.
(a) A faucet on the second floor, 7.0 m above the first floor, is turned off. What is the gauge pressure at this faucet?




P2= P1 + (density x g x h)



(2 x 10^5)= P1 + (1000 x 9.81 x 7)

P1= 131330 Pa
 
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Looks OK.
 
Ah, I see the error. Why is the pipe entering the house on the first floor? We all know pipes enter through the basement. Therefore, the height should be negative.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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