A sniper fires a rifle bullet into a gasoline | vedclass

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(B) According to Bernoulli's theorem at points $A$ (the hole) and $B$ (the surface of the gasoline):$P_B + \rho gh = P_A + \frac{1}{2}\rho v_A^2$Given

Vedclass · Accepted Answer

$41.0$

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(B) According to Bernoulli's theorem at points $A$ (the hole) and $B$ (the surface of the gasoline):$P_B + ho gh = P_A + \frac{1}{2}ho v_A^2$Given:$P_B = 3.10 \ atm = 3.10 	imes 1.013 	imes 10^5 \ Pa$$P_A = 1.00 \ atm = 1.013 	imes 10^5 \ Pa$$h = 53.0 \ m$$ho = 660 \ kg \ m^{-3}$$g = 9.8 \ m \ s^{-2}$Substituting the values:$3.10 	imes 1.013 	imes 10^5 + 660 	imes 9.8 	imes 53.0 = 1.013 	imes 10^5 + \frac{1}{2} 	imes 660 	imes v_A^2$$(3.10 - 1.00) 	imes 1.013 	imes 10^5 + 342888 = 330 	imes v_A^2$$2.10 	imes 1.013 	imes 10^5 + 342888 = 330 	imes v_A^2$$212730 + 342888 = 330 	imes v_A^2$$555618 = 330 	imes v_A^2$$v_A^2 = \frac{555618}{330} \approx 1683.69$$v_A = \sqrt{1683.69} \approx 41.0 \ m \ s^{-1}$

$A$ sniper fires a rifle bullet into a gasoline tank,making a hole $53.0 \ m$ below the surface of the gasoline. The tank was sealed at $3.10 \ atm$. The stored gasoline has a density of $660 \ kg \ m^{-3}$. The velocity with which gasoline begins to shoot out of the hole is ........ $m \ s^{-1}$.

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