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09:40:01 query introset plug of water from cylinder given gauge pressure, c-s hole area, length of plug.
Explain how we can determine the velocity of exiting water, given the pressure difference between inside and outside, by considering a plug of known cross-sectional area and length?......!!!!!!!!...................................
RESPONSE --> The net force on the plug is P * A. L is the length of the plug. It is doing work of P*A*L. The volume of the plug is A*L so the mass is RHO*A*L. If we have .5mv^2=KE we have .5rhoA L v^2= P*A*L= v=sqrt(2P/rho).
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09:41:11 ** 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 P * A acts thru dist L doing work P * A * L. If init vel 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). **......!!!!!!!!...................................
RESPONSE --> ok i understand
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09:45:47 query gen phy 10.24 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force
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RESPONSE --> The volume of the balloon is 4/3 pi r^3 = 1662 cubic meters mass of air displaced= 1.3 kg / m^3 * 1662 m^3 = 2161 kg. The buoyant force is equal in magnitude to the force of gravity on the displaced airo So it is 2161 kg * 9.8 m/s^2 =21178 N. buoyant force - weight = 21,178 N - 9100 N = 12,078 N
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09:48:19 ** 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 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.. **......!!!!!!!!...................................
RESPONSE --> my numbers are slightly different. Also I did not find the net force when the helium is accounted for. So at .18kg/m^3 the 1660 m^3 of helium will have a mass of 300kg on which the gravity exerts a force of 2900 N. The net force would be 11,400 N-2900N= 8,500 N. So the force = 8500N/(9.8m/s^2)=870kg.
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09:49:26 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.
Give your solution to this problem.......!!!!!!!!...................................
RESPONSE --> DNA
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09:50:33 ** At the interface the pressure is that of the atmosphere plus 15 cm of water, or 1 atm + 1000 kg/m^3 * 9.8 m/s^2 * .15 m = 1 atm + 1470 Pa.
The 15 cm of water above the water-mercury interface must be balanced by the mercury above this level on the other side. Since mercury is 13.6 times denser than water there must therefore be 15 cm / (13.6) = 1.1 cm of mercury above this level on the other side. This leaves the top of the water column at 15 cm - 1.1 cm = 13.9 cm above the mercury-air interface. } Here's an alternative solution using Bernoulli's equation, somewhat more rigorous and giving a broader context to the solution: Comparing the interface between mercury and atmosphere with the interface between mercury and water we see that the pressure difference is 1470 Pa. Since velocity is zero we have P1 + rho g y1 = P2 + rho g y2, or rho g (y1 - y2) = P2 - P1 = 1470 Pa. Thus altitude difference between these two points is y1 - y2 = 1470 Pa / (rho g) = 1470 Pa / (13600 kg/m^2 * 9.8 m/s^2) = .011 m approx. or about 1.1 cm. The top of the mercury column exposed to air is thus 1.1 cm higher than the water-mercury interface. Since there are 15 cm of water above this interface the top of the water column is 15 cm - 1.1 cm = 13.9 cm higher than the top of the mercury column. NOTE BRIEF SOLN BY STUDENT: Using Bernoullis Equation we come to: 'rho*g*y1='rho*g*y2 1*10^3*9.8*.15 =13.6*10^3*9.8*y2 y2=.011 m h=y1-y2 h=.15-.011=.139m h=13.9cm. **......!!!!!!!!...................................
RESPONSE --> DNA
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