Water is moving with a speed of 5.0,m/s throu | vedclass

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(B) According to the equation of continuity,$A_1 v_1 = A_2 v_2$.Given $A_1 = 4.0\,cm^2$,$v_1 = 5.0\,m/s$,and $A_2 = 8.0\,cm^2$.$v_2 = (A_1 v_1) / A_2

Vedclass · Accepted Answer

$2.7 	imes 10^5\,Pa$

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(B) According to the equation of continuity,$A_1 v_1 = A_2 v_2$.Given $A_1 = 4.0\,cm^2$,$v_1 = 5.0\,m/s$,and $A_2 = 8.0\,cm^2$.$v_2 = (A_1 v_1) / A_2 = (4.0 	imes 5.0) / 8.0 = 2.5\,m/s$.According to Bernoulli's theorem,$P_1 + ho g h_1 + 1/2 ho v_1^2 = P_2 + ho g h_2 + 1/2 ho v_2^2$.Rearranging for $P_2$: $P_2 = P_1 + ho g(h_1 - h_2) + 1/2 ho(v_1^2 - v_2^2)$.Here,$h_1 - h_2 = 10\,m$,$ho = 1000\,kg/m^3$,$g = 10\,m/s^2$.$P_2 = 1.5 	imes 10^5 + (1000 	imes 10 	imes 10) + 0.5 	imes 1000 	imes (5^2 - 2.5^2)$.$P_2 = 1.5 	imes 10^5 + 1.0 	imes 10^5 + 500 	imes (25 - 6.25)$.$P_2 = 2.5 	imes 10^5 + 500 	imes 18.75 = 2.5 	imes 10^5 + 9375 = 2.59375 	imes 10^5\,Pa \approx 2.7 	imes 10^5\,Pa$.

Water is moving with a speed of $5.0\,m/s$ through a pipe of cross-sectional area $4.0\,cm^2$. The water gradually descends $10\,m$ as the pipe increases in area to $8.0\,cm^2$. If the pressure at the upper level is $1.5 \times 10^5\,Pa$,the pressure at the lower level will be:

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