course Phy 202
.................................................??????K?y^???????
......!!!!!!!!...................................
00:43:02 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 --> Force=P (in N/m^2)*a (in m^2), so the meters cancel out leaving only Newtons of force. Now, work will be the number derived above, the force, multiplied by the length L. For the volume, we multiply the cross sectional area by the length. For the mass, we multiply the density by the volume. Finally, we set the KE=P*V, .5mv^2=PV, and solve for little v, the velocity.
.................................................
......!!!!!!!!...................................
00:43:08 ** 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 follow this procedure well.
.................................................
......!!!!!!!!...................................
00:51:21 prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m.
......!!!!!!!!...................................
RESPONSE --> The volume is 4.8m*3.8m*2.8m=about 51m^3 The average room air density is about 1.3 kg/m^3 Now i can use the formula D=m/v, so mass=d*v= 1.3 kg / m^3*51 m^3=66.3 kg
.................................................
......!!!!!!!!...................................
00:51:24 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
.................................................
......!!!!!!!!...................................
00:54:32 prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.
......!!!!!!!!...................................
RESPONSE --> I don't know what the density of blood is, so I am going to do a hypothetical situation synposis of the question instead of working it out explicitly. First, I have to know the density, gravity, and the length in order to find the pressure, for which I just multiply together all these quantities. Then, well, I am sort of lost. I thought I knew what to do but I don't.
.................................................
......!!!!!!!!...................................
00:58:38 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 see--blood and water are similar enough that, for our purposes, they can basically be interchanged. I actually didn't realize or had forgotten, since I didn't actually do the calculations, that I would need to convert N/m^2 to Pascals. Then, I see that I just have to divide the Pascals of pressure by the pressure of 1 mm of mercury, which, you explained, is how blood pressure is measured. The division gave a number 117 mm of mercury which is the difference between head and toe.
.................................................
......!!!!!!!!...................................
05:42:57 prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force
......!!!!!!!!...................................
RESPONSE --> Assuming the balloon is close enough to spherical, I can use 4/3 pi r^2 to calculate the volume. 4/3 pi (7.35)^3=1660 m^3 Mass of the displaced air is its density * volume, or 1.3 kg/m^3*1660 m^3=2160 kg. Now buoyant force, differing from net force, calculates on the force of gravity as its acts on the air that is displaced. So, we have 2160 kg*9.8 m/s^2=21168 N. Now 930 kg*9.8m/s^2=9114 N. THe net force would then be 21168 N-9114N=12054N
.................................................
......!!!!!!!!...................................
05:51:38 ** 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 that I should have also taken into account the mass of the helium itself and should have substructed the prodcut of helium's mass and the force of gravity from the Net force acting on the balloon.
.................................................
......!!!!!!!!...................................
05:51:40 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 -->
.................................................
......!!!!!!!!...................................
05:51:41 ** 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 -->
................................................."