course Phys 202 005. `query 5
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Given Solution: ** The net force on the plug is P * A, where A is its cross-sectional area and P the pressure difference. If L is the length of the plug then the net force F_net = P * A acts thru distance L doing work `dW = F_net * L = P * A * L. If the initial velocity of the plug is 0 and there are no dissipative forces, this is the kinetic energy attained by the plug. The volume of the plug is A * L so its mass is rho * A * L. Thus we have mass rho * A * L with KE equal to P * A * L. Setting .5 m v^2 = KE we have .5 rho A L v^2 = P A L so that v = sqrt( 2 P / rho). Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m. ********************************************* Question: prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person. ********************************************* Question: prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Fb = (mHe + mcargo)g rho air * V*g = (mHe + mcargo)g 1.29 kg/m^3 * (4/3*pi*(7.35m)^2)) * 9.8 m/s^2 = 930 kg + mcargo 21000 N = (930 kg + mcargo) (9.8 m/s^2) 2150 = 930 kg + mcargo 1200 kg = mcargo Confidence Rating: 3
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Given Solution: ** The volume of the balloon is about 4/3 pi r^3 = 1660 cubic meters and mass of air displaced is about 1.3 kg / m^3 * 1660 m^3 = 2160 kg. The buoyant force is equal in magnitude to the force of gravity on the displaced air, or about 2160 kg * 9.8 m/s^2 = 20500 Newtons, approx.. If the total mass of the balloon, including helium, is 930 kg then the net force is about Net force = buoyant force - weight = 20,500 N - 9100 N = 11,400 N If the 930 kg doesn't include the helium we have to account also for the force of gravity on its mass. At about .18 kg/m^3 the 1660 m^3 of helium will have mass about 300 kg on which gravity exerts an approximate force of 2900 N, so the net force on the balloon would be around 11,400 N - 2900 N = 8500 N approx. The mass that can be supported by this force is m = F / g = 8500 N / (9.8 m/s^2) = 870 kg, approx.. ** Your Self-Critique: I forgot to take out the mass of the Helium and the force of gravity on its mass. rho * V = m 0.18 kg/m^3 * 1660 m^3 = 300 kg If I subtract out those 300 kg I end up with about 915 kg. My answer is different because I rounded less. Your Self-Critique Rating: 3 ********************************************* Question: univ 14.51 (14.55 10th edition) U tube with Hg, 15 cm water added, pressure at interface, vert separation of top of water and top of Hg where exposed to atmosphere.