A horizontal section of an artery narrows from a diameter of 8.0 mm to a diameter of 6.0 mm.What is the pressure difference between the wider section and the narrower? That is,what is P_wide - P_narrow? The blood flows in the wider section at 48 cm/s.Take blood plasma to have a density of 1025 kg/m³ and ignore viscosity. A) 370 Pa B) 750 Pa C) 780 Pa D) 210 Pa

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The pressure difference is given by the Bernoulli equation:$$P_1+\frac{1}{2}ho v_1^2=P_2+\frac{1}{2}ho v_2^2$$where $ho$ is the density of the fluid, $v$ is the velocity of the fluid, and $P$ is the pressure of the fluid.In this case, the fluid is blood plasma, the velocity of the fluid in the wider section is 48 cm/s, and the velocity of the fluid in the narrower section is unknown.We can use the equation of continuity to find the velocity of the fluid in the narrower section:$$A_1v_1=A_2v_2$$where $A$ is the cross-sectional area of the pipe.In this case, the cross-sectional area of the wider section is $\pi(8.0	ext{ mm}/2)^2=50.3	ext{ mm}^2$, and the cross-sectional area of the narrower section is $\pi(6.0	ext{ mm}/2)^2=28.3	ext{ mm}^2$.Therefore, the velocity of the fluid in the narrower section is:$$v_2=\frac{A_1v_1}{A_2}=\frac{50.3	ext{ mm}^2	imes48	ext{ cm/s}}{28.3	ext{ mm}^2}=84	ext{ cm/s}$$Now we can use the Bernoulli equation to find the pressure difference between the wider section and the narrower section:$$P_1+\frac{1}{2}ho v_1^2=P_2+\frac{1}{2}ho v_2^2$$$$P_1-P_2=\frac{1}{2}ho(v_2^2-v_1^2)$$$$P_1-P_2=\frac{1}{2}	imes1025	ext{ kg/m}^3	imes(84	ext{ cm/s})^2-(48	ext{ cm/s})^2$$$$P_1-P_2=370	ext{ Pa}$$Therefore, the pressure difference between the wider section and the narrower section is 370 Pa.

A Horizontal Section of an Artery Narrows from a Diameter