Question

Please help with how to find the answers for part a and b! Use the data describing blood flow in the circulatory system from the table below, and assume a typical blood flow rate of 5.o I per minute AortaArteries Arterioles Capillaries Venules Veins Vena cava 2 0.5 0.002 0.0009 0.003 0.5 3 3 20 500 4000 3000 80 7 Diameter (cm) Total area (cm2) Part A What is the flow speed in the arterioles? Express your answer with the appropriate units. U = 1.7x10-3 m s Submit Previous Answers Correct Part B What is the pressure difference across a 2.0 cm length of arteriole? Express your answer with the appropriate units. C ] p = 1.6 103 Pa Submit Previous Answers Request Answer XIncorrect; Try Again;8 attempts remaining

Answer 1

The flow speed in the arterioles is 2500 cm/s, and the pressure difference across a 2.0 cm length of arteriole is 2.08 x 10^5 dyn/cm² or 2.08 x 10^3 Pa.

Step-by-step explanation:

To find the flow speed in the arterioles, we can use the equation Q = Au, where Q is the flow rate, A is the cross-sectional area, and u is the flow speed. From the table, the cross-sectional area of the arterioles is 0.002 cm². We know that the flow rate is 5.0 L/min, which is equivalent to 5.0 cm³/s. Rearranging the equation, we get u = Q/A. Plugging in the values, we have u = 5.0 cm³/s / 0.002 cm² = 2500 cm/s.

To find the pressure difference across a 2.0 cm length of arteriole, we can use the equation Δp = F/A, where Δp is the pressure difference, F is the force, and A is the cross-sectional area. The force can be calculated using the equation F = ma, where m is the mass and a is the acceleration. Assuming a typical blood flow rate of 5.0 L/min, the mass of the blood passing through the arteriole in 2.0 cm length is 5.0 cm³/s × 1 g/cm³ × 2.0 cm = 10 g. The acceleration can be calculated using the equation a = Δv/t, where Δv is the change in velocity and t is the time. Assuming a change in velocity of 2500 cm/s (as calculated previously) over a time of 1 minute (60 seconds), we have a = 2500 cm/s / 60 s = 41.67 cm/s². Plugging these values into the equation F = ma, we get F = 10 g × 41.67 cm/s² = 416.7 g·cm/s² = 416.7 dyn. The cross-sectional area of the arteriole is 0.002 cm² (as calculated previously). Plugging these values into the equation Δp = F/A, we have Δp = 416.7 dyn / 0.002 cm² = 2.08 x 10^5 dyn/cm² or 2.08 x 10^3 Pa.

Answer 2

To find the flow speed in the arterioles, use the equation Q = Au and substitute the values from the table.

Step-by-step explanation:

To find the flow speed in the arterioles, we can use the equation Q = Au, where Q is the flow rate, A is the cross-sectional area, and u is the flow speed. From the table, we can see that the total area of the arterioles is 3 cm². We can substitute these values into the equation to solve for u:

Q = Au
5 L/min = 3 cm² * u
u = 5 L/min ÷ 3 cm²
u ≈ 1.7x10-3 cm/min