Question 1 What hanging mass will stretch a 1.5-m-long, 0.64mm - diameter steel
ID: 2014056 • Letter: Q
Question
Question 1What hanging mass will stretch a 1.5-m-long, 0.64mm - diameter steel wire by 1.0mm?
Question 2
The density of aluminum is 2700 kg/m^3. How many atoms are in a 1.20cm x 1.20cm x 1.20cm cube of aluminum?
Question 3
Smoking tobacco is bad for your circulatory health. In an attempt to maintain the blood's capacity to deliver oxygen, the body increases its red blood cell production, and this increases the viscosity of the blood. In addition, nicotine from tobacco causes arteries to constrict. For a nonsmoker, with blood viscosity of 2.5x10^-3 Pa*s, normal blood flow requires a pressure difference of 8.0 mm of Hg between the two ends of an artery. If this person were to smoke regularly, his blood viscosity would increase to 2.7 x10^-3 Pa*s, and the arterial diameter would constrict to 90% of its normal value. What pressure difference would be needed to maintain the same blood flow?
Question 4
A stiff, 15 cm-long tube with an inner diameter of 3.0 mm is attached to a small hole in the side of a tall beaker. The tube sticks out horizontally. The beaker is filled with 20 degrees Celsius water to a level 50 cm above the hole, and it is continually topped off to it maintain that level.What is the volume flow rate through the tube?
Explanation / Answer
1) Length of the wire L = 1.50 m Elongation in the wire L = 1.0 mm = 1.0*10^-3 m Diameter of the wire d = 0.64 mm Radius of the wire r = d/2 = 0.32 mm = 0.32*10^-3 m Cross sectional area of the wire A = r^2 = (0.32*10^-3 m)^2 = 3.216*10^-7 m^2 Young's modulus of steel Y = 2*10^11 N/m^2 Let m be the required mass. Formula for Youngs modulus Y = mgL/(L)A Mass m = YA(L)/gL = (2*10^11 N/m^2)(3.216*10^-7 m^2)(1.0*10^-3 m)/(9.8 m/s^2)(1.50m) = 4.375 Kg (2) Density of Aluminum = 2700 Kg/m^3 Volume of the cube V = (1.2 cm)^3 = 1.728 cm^3 = 1.728*10^-6 m^3 Number density d = /atomic mass = (2700 Kg/m^3)/(26.98) = 100.07 Kg/m^3 Number of moles n = d * V = (100.07 Kg/m^3) * ( 1.728*10^-6 m^3) = 1.729*10^-4 mol Number of atoms = n * Avagadro number = (1.729*10^-4 mol) * (6.0221415 × 10^23) = 1.04*10^20 Note : Our rules does not allow us to answer more than one question in a single post.So please post the remaining seperately. Thank you so much. Length of the wire L = 1.50 m Elongation in the wire L = 1.0 mm = 1.0*10^-3 m Diameter of the wire d = 0.64 mm Radius of the wire r = d/2 = 0.32 mm = 0.32*10^-3 m Cross sectional area of the wire A = r^2 = (0.32*10^-3 m)^2 = 3.216*10^-7 m^2 Young's modulus of steel Y = 2*10^11 N/m^2 Let m be the required mass. Formula for Youngs modulus Y = mgL/(L)A Mass m = YA(L)/gL = (2*10^11 N/m^2)(3.216*10^-7 m^2)(1.0*10^-3 m)/(9.8 m/s^2)(1.50m) = 4.375 Kg (2) Density of Aluminum = 2700 Kg/m^3 Volume of the cube V = (1.2 cm)^3 = 1.728 cm^3 = 1.728*10^-6 m^3 Number density d = /atomic mass = (2700 Kg/m^3)/(26.98) = 100.07 Kg/m^3 Number of moles n = d * V = (100.07 Kg/m^3) * ( 1.728*10^-6 m^3) = 1.729*10^-4 mol Number of atoms = n * Avagadro number = (1.729*10^-4 mol) * (6.0221415 × 10^23) = 1.04*10^20 Note : Our rules does not allow us to answer more than one question in a single post.So please post the remaining seperately. Thank you so much.Related Questions
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