assignment #004 rYɐZ Physics II 03-06-2006
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22:34:31 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 --> This would be the KE of the flow through the plug. The work = force * distance = ( `dP * A) * L
The mass of the plug is m = `rho * V = `rho * L * A. Substituing that into KE =.5mv^2 we get the formula .5 m v^2 = `dP * A * L, or .5 `rho * L * A * v^2 = `dP * A * L. When we solve for v we use v = `sqrt( 2 * dP * A * L / (`rho * L * A) ) = `sqrt ( 2 `dP / `rho)..................................................
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22:37:06 ** 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
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22:37:10 prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m.
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RESPONSE --> ok
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22:37:14 The density of air at common atmospheric temperature and pressure it about 1.3 kg / m^3.
The volume of the room is 4.8 m * 3.8 m * 2.8 m = 51 m^3, approximately. The mass of the air in this room is therefore mass = density * volume = 1.3 kg / m^3 * 51 m^3 = 66 kg, approximately. This is a medium-sized room, and the mass is close to the average mass of a medium-sized person.......!!!!!!!!...................................
RESPONSE --> ok
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22:37:21 prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.
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RESPONSE --> ok
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22:37:25 The density of blood is close to that of water, 1000 kg / m^3. So the pressure difference corresponding to a height difference of 1.6 m is
pressure difference = rho g h = 1000 kg/m^3 * 9.80 m/s^2 * 1.60 m = 15600 kg / ( m s^2) = 15600 (kg m / s^2) / m^2 = 15600 N / m^2, or 15600 Pascals. 1 mm of mercury is 133 Pascals, so 15,600 Pa = 15,600 Pa * ( 1 mm of Hg / 133 Pa) = 117 mm of mercury. Blood pressures are measured in mm of mercury; e.g. a blood pressure of 120 / 70 stands for a systolic pressure of 120 mm of mercury and a diastolic pressure of 70 mm of mercury.......!!!!!!!!...................................
RESPONSE --> ok
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22:44:58 prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force
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RESPONSE --> We know that the density of air is 1.29kg/m^3 and the density of helium is .179kg/m^3.
First we must find the volume of the baloon v=4/3'pi r^3 v=4/3 'pi 7.35^3=1663.22 We must know find the force Fb=1.29*1663.22*9.8m/s=21026.4N It tells us that the total weight of the baloon is 930kg and we know that 1kg =9.8N so 930*9.8=9114N Now we just subtract the two F=21026.4-9114=11912N We must now consider the helium which is less dense than air and will aid in lifting up the balloon. .179*1663.22*9.8=2917.62N We know just subtract the helium calculations from our previous. 11912N-2917.62=8994.38N is the buoyant force.................................................
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22:47:16 ** 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 --> I failed to convert it back to kg. We just rearranged F=M*A to accomplish this. m=F/g=8994.38/9.8=917.8kg
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22:47:19 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 --> ok
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