Air streams horizontally past an airplane. Th | vedclass

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(A) According to Bernoulli's principle for horizontal flow $(h_1 = h_2)$: $P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$ where $P_1$ an

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$1018$

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(A) According to Bernoulli's principle for horizontal flow $(h_1 = h_2)$: $P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho v_2^2$ where $P_1$ and $v_1$ are pressure and velocity at the top surface,and $P_2$ and $v_2$ are pressure and velocity at the bottom surface. The pressure difference $\Delta P = P_2 - P_1$ is given by: $\Delta P = \frac{1}{2} ho (v_1^2 - v_2^2)$ Given: $ho = 1.293 \, kg/m^3$,$v_1 = 60 \, m/s$,$v_2 = 45 \, m/s$. $\Delta P = \frac{1}{2} 	imes 1.293 	imes (60^2 - 45^2)$ $\Delta P = 0.6465 	imes (3600 - 2025)$ $\Delta P = 0.6465 	imes 1575$ $\Delta P \approx 1018.23 \, N/m^2$. Thus,the difference in pressure is approximately $1018 \, N/m^2$.

Air streams horizontally past an airplane. The speed over the top surface is $60 \, m/s$ and that under the bottom surface is $45 \, m/s$. The density of air is $1.293 \, kg/m^3$. The difference in pressure is ....... $N/m^2$.

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