course PHY 202 ZSݎyassignment #014
......!!!!!!!!...................................
20:17:17 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 --> When asked to calculate the velocity of exiting water, it is very useful to use Work to obtain it, where W=FD, where F is Force and D is the displacement of the water(or length in this case). Force is equal to Area(Gauge Pressure), where gauge pressure is the difference between the inside and outside. If given area and gauge pressure, it is an easy calculation for Force. Then stick Force into W=FL to find Work. Once we know work, we can use the W=.5mv^2 to solve for the velocity.
.................................................
......!!!!!!!!...................................
20:19:42 ** 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 --> Forgot to explain how to obtain Mass for the formula W=.5mv""2. To obtain mass, you can use the formula rho=m/v, where rho=density, m=mass and v=volume. Volume is obtained by L(A), where l=length and A=area.
.................................................
......!!!!!!!!...................................
20:34:07 prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force
......!!!!!!!!...................................
RESPONSE --> Given the formula: Buoyant Force=m(of fluid displaced)g; where m= mass of fluid displaced, and g=acceleration due to gravity. It follows that 930(9.80)=9114 Newtons. We are given radius of the balloon. In most cases, if I were not given the mass, I would use the radius to find the volume, which in this case is 4/3(pi)(radius^3). Then I would multiply the volume times the density to get mass.
.................................................
......!!!!!!!!...................................
20:52:11 ** 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 --> So I did need to use the radius to find the volume and multiply it by the density of AIR to get the mass of the displaced AIR. I must remember that Buoyant Force on an object immersed in fluid(in this case air) is equal to the weight of the fluid displaced by that object. So it's 2160(9.8),. Then Fb=F1-F2, where Fb=Buoyant Force, F1=Force on air, F2=Force on helium balloon, because net force has to equal zero.
.................................................
~z䘻 assignment #014 014. `query 4 Physics II 07-05-2007"