course phy 122 014. `query 4Physics II
......!!!!!!!!...................................
13:09:54 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 --> we can use the equation P*A *Lto come up with the force P is the difference in the pressure and A is the cross-sectional area, and L is the length of the plug im not sure how to come up with the rest of the solution...
.................................................
......!!!!!!!!...................................
13:10:55 ** 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 --> I understand how to find the volume of the plug and how to also find the mass with the equations.
.................................................
......!!!!!!!!...................................
13:13:22 prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m.
......!!!!!!!!...................................
RESPONSE --> we can easily fnd the volume with 4.8 m * 3.8 m * 2.8 m= 51 m^3 so we can then find the mass by, density*volume
.................................................
......!!!!!!!!...................................
13:14:33 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 --> I understand that the density of the air at common atmospheric temperature and pressure is about 1.3 kg/m^3, and this would be plugged into the equatin with the volume to find the mass.
.................................................
......!!!!!!!!...................................
13:17:21 prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.
......!!!!!!!!...................................
RESPONSE --> the pressure difference is equal to rho g h i looked up the density of blood, and saw that it was similar to water, which is 100 kg / m^3 pressure difference would be 1.8 m 1000 kg/ m^3 * 9.80 m/s^2* 1.6 m= 15600 kg / (m s^2)
.................................................
......!!!!!!!!...................................
13:17:52 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 understand that blood pressures are measured in mm of mercury, but i did not realize that until reading this.
.................................................
......!!!!!!!!...................................
13:18:51 prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force
......!!!!!!!!...................................
RESPONSE --> im not very sure
.................................................
......!!!!!!!!...................................
13:21:56 ** 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 understand how to find the volume of the balloon and that the buoyant force is equal in magnitude to the force of gravity on the displaced air
.................................................
......!!!!!!!!...................................
13:22:04 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 --> this is for university students
.................................................
......!!!!!!!!...................................
13:22:06 ** 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 --> okay