course PHY122 ????b???????assignment #014
......!!!!!!!!...................................
13:46:37 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 pressure is gauge pressure, so the reading on the gauge is the pressure difference betwen the outside and the inside temperature. You can find the force by multiplying the area * the pressure. This will give you an answer in N The force * the distance will give you the amount of work required. Befor we can find the mass we need to find the volume of the plug. witch is The area*the length. Once we have the mass we can then put it KE formula to get is's velocity. KE= .5m*v^2 Using the amount of work perfomed by the plug and the KE formula we can sovle for velocity.
.................................................
......!!!!!!!!...................................
13:47:00 ** 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
.................................................
......!!!!!!!!...................................
13:56:16 prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m.
......!!!!!!!!...................................
RESPONSE --> The dinsity of air is 1.29kg/m^3 The voume of air is 4.8m*3.8m*2.8m=51.072m^3 Mass = volume * density 51.072m^3*1.29kg/m^3= 65.88kg Mass = 65.88kg
.................................................
......!!!!!!!!...................................
13:57:06 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
.................................................
......!!!!!!!!...................................
14:16:56 prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.
......!!!!!!!!...................................
RESPONSE --> d'P=density*accelaration due to gravity*d'h Density of blood =1.05*10^3kg/m^3 Since the height at the feet would be 0, and the height at the head would be 1.6m The pressure at the head would be the difference. (1.05*10^3kg/m^3) 9.8m/s^2*1.6m=16464n/m^2
.................................................
......!!!!!!!!...................................
14:18:41 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 --> I used the density for blood in the book and didn't do the mm of mecury part the you did, but my answer was close.
.................................................
......!!!!!!!!...................................
14:58:46 prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force
......!!!!!!!!...................................
RESPONSE --> We know the mass of helium, 930kg, so to find it's weight we multiply the m*g and get 930kg*9.8m/s^2 and get 9114N To get the mass of the air we find the volume of the ballon by 4/3pi'*r^3 4/3*3.14*7.35^3=1663.22m^3 m=V*density; 1663.22m^3*1.29kg/m^3=2145.5kg so the weight of the displaced air is 2145.5kg*9.8m/s^2=21025N The difference in these two would be the bouyant force 2145.5N-9114N=6968.5N
.................................................
......!!!!!!!!...................................
15:01:14 ** 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 got half of the answer, I just didn't figure the mass that could be supported by the ballon.
.................................................
"