22. Suppose your mass is 70.0 kg and your density is 970kg/m^3 If you could stand on a scale in a vacuum chamber on Earth's surface the reading of the scale would be mg=(170.0kg)(9.80N/kg^2)=686 N. What will the scale read when you are completely submerged in air of density 1.29kg/m^3 (b) What will the scale read if you weigh yourself in a swimming pool with your body completely submerged? 23. The aorta is the principal blood vessel through which blood leaves the heart in order to circulate around the body. (a) Calculate the average speed of the blood in the aorta if the flow rate is 5.0L/min The aorta has a radius of 10mm. (b)Blood also flows through smaller blood vessels known as capillaries. When the rate of blood flow in the aorta is 5.0L/min the speed of blood in the capillaries is about 0.33mm/s. Given that the average diameter of a capillary is 8.0mu m calculate the number of capillaries in the blood circulatory system.

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22. Suppose your mass is 70.0 kg and your density is 970kg/m^3 If you could stand on a scale in a vacuum chamber on Earth's surface the reading of the scale would be mg=(170.0kg)(9.80N/kg^2)=686 N. What will the scale read when you are completely submerged in air of density 1.29kg/m^3 (b) What will the scale read if you weigh yourself in a swimming pool with your body completely submerged? 23. The aorta is the principal blood vessel through which blood leaves the heart in order to circulate around the body. (a) Calculate the average speed of the blood in the aorta if the flow rate is 5.0L/min The aorta has a radius of 10mm. (b)Blood also flows through smaller blood vessels known as capillaries. When the rate of blood flow in the aorta is 5.0L/min the speed of blood in the capillaries is about 0.33mm/s. Given that the average diameter of a capillary is 8.0mu m calculate the number of capillaries in the blood circulatory system.

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# Brief Explanations: For question 22, we need to calculate the buoyant force exerted by air and water on a person and determine the reading on the scale. For question 23, we need to use the flow rate and the cross-sectional area to find the speed of blood in the aorta and the number of capillaries. # Answer: ### Question 22: (a) The buoyant force in air can be calculated using the density of air and the volume of the person. The volume \( V \) of the person can be found using their mass and density. \[ V = \frac{m}{\rho} = \frac{70.0 \, \text{kg}}{970 \, \text{kg/m}^3} = 0.0722 \, \text{m}^3 \] The buoyant force \( F_b \) in air is: \[ F_b = \rho_{\text{air}} \cdot V \cdot g = 1.29 \, \text{kg/m}^3 \cdot 0.0722 \, \text{m}^3 \cdot 9.80 \, \text{N/kg} = 0.913 \, \text{N} \] The scale reading in air is: \[ F_{\text{scale}} = mg - F_b = 686 \, \text{N} - 0.913 \, \text{N} = 685.087 \, \text{N} \] (b) The buoyant force in water is: \[ F_b = \rho_{\text{water}} \cdot V \cdot g = 1000 \, \text{kg/m}^3 \cdot 0.0722 \, \text{m}^3 \cdot 9.80 \, \text{N/kg} = 707.56 \, \text{N} \] The scale reading in water is: \[ F_{\text{scale}} = mg - F_b = 686 \, \text{N} - 707.56 \, \text{N} = -21.56 \, \text{N} \] Since the reading cannot be negative, it indicates that the person would float, and the scale would read zero. ### Question 23: (a) The flow rate \( Q \) is given by: \[ Q = A \cdot v \] Where \( A \) is the cross-sectional area of the aorta and \( v \) is the average speed of the blood. The area \( A \) is: \[ A = \pi r^2 = \pi (0.01 \, \text{m})^2 = 3.14 \times 10^{-4} \, \text{m}^2 \] The flow rate \( Q \) is: \[ Q = 5.0 \, \text{L/min} = 5.0 \times 10^{-3} \, \text{m}^3/\text{min} = 8.33 \times 10^{-5} \, \text{m}^3/\text{s} \] The average speed \( v \) is: \[ v = \frac{Q}{A} = \frac{8.33 \times 10^{-5} \, \text{m}^3/\text{s}}{3.14 \times 10^{-4} \, \text{m}^2} = 0.265 \, \text{m/s} \] (b) The flow rate in capillaries is the same as in the aorta. The cross-sectional area of a capillary is: \[ A_{\text{capillary}} = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{8.0 \times 10^{-6} \, \text{m}}{2}\right)^2 = 5.027 \times 10^{-11} \, \text{m}^2 \] The number of capillaries \( N \) is: \[ N = \frac{Q}{A_{\text{capillary}} \cdot v_{\text{capillary}}} = \frac{8.33 \times 10^{-5} \, \text{m}^3/\text{s}}{5.027 \times 10^{-11} \, \text{m}^2 \cdot 0.33 \times 10^{-3} \, \text{m/s}} = 5.0 \times 10^9 \] # Answer: ### Question 22: (a) 685.087 N (b) 0 N (indicating floatation) ### Question 23: (a) 0.265 m/s (b) 5.0 × 10^9 capillaries

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