#$&* course PHY 202 005. `query 5 was submitted 11 Jun 2011 around 9:30 PM. 005. `query 5*********************************************
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Given Solution: ** 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 F_net = P * A acts thru distance L doing work `dW = F_net * L = P * A * L. If the initial velocity of the plug 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). STUDENT SOLUTION AND QUESTION From looking at my notes from the Intro Problem Set, it was a lot of steps to determine the velocity in that situation. The Force was determined first by using F = (P * cross-sectional area). With that obtained Force, use the work formula ‘dW = F * ‘ds using the given length as the ‘ds. Then we had to obtain the Volume of the ‘plug’ by cross-sectional area * length. That Volume will then be using to determine the mass by using m = V * d, and the density is a given as 1000kg/m^3. With that mass, use the KE equation 1/2(m * v^2) to solve for v using the previously obtained ‘dW for KE since ‘dW = ‘dKE. Some of my symbols are different than yours. I’m in your PHY 201 class right now and am in the work and energy sections. So is it wrong to use 1/2(m * v^2) instead of 1/2(‘rho * A * L * v^2)?? Aren’t they basically the same thing?? INSTRUCTOR RESPONSE You explained the process very well, though you did miss a step. m = rho * V (much better to use rho than d, which isn't really a good letter to use for density or distance, given its use to represent the prefix 'change in'). You covered this in your explanation. V isn't a given quantity; it's equal to the length of the plug multiplied by its cross-sectional area and should be expressed as such. You didn't cover this in your explanation. However, other than this one missing step, your explanation did cover the entire process. With that one additional step, your solution would be a good one. In any case, V would be A * L, and m would be rho * V = rho * A * L. So 1/2 m v^2 is the same as 1/2 rho A L v^2. Your Self-Critique: ok Your Self-Critique Rating: 3 ********************************************* Question: prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The density of air at common atmospheric temperature and pressure it about 1.3 kg / m^3. room’s volume = 4.8 m * 3.8 m * 2.8 m = 51 m^3 The mass of the air in this room is therefore: m = d * v m= 1.3 kg / m^3 * 51 m^3 = 66.3 kg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 of the air in that room is close to the mass of an average-sized person. Your Self-Critique: ok Your Self-Critique Rating: 3 ********************************************* Question: prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The volume of the balloon is approx 4/3 pi r^3 = 1660 m^3. The mass of air displaced is approx 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 which is approx: B (f) = 2160 kg * 9.8 m/s^2 B (f) = 20500 N If the total mass of the balloon, including helium, is 930 kg then the weight is: 930 kg * 9.8 m/s^2 = 9100 N F_net = B (f) - wt F_net = 20,500 N - 9100 N F_net = 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 m = 8500 N / (9.8 m/s^2) = 870 kg STUDENT QUESTION I got part of the problem right. I don’t understand the volume of air displaced….. INSTRUCTOR RESPONSE The 1660 m^3 volume of the balloon takes up 1660 m^3 that would otherwise be occupied by the surrounding air. The surrounding air would be supporting the weight of the displaced air, if the balloon wasn't there displacing it. That is, the surrounding air would act to support 20500 Newtons of air. The surrounding air is no different for the fact something else is there, instead of the air displaced air. So it supports 20500 Newtons of whatever is there displacing the air it would otherwise be supporting. This supporting force of 20500 Newtons is therefore exerted on the balloon. We call this the buoyant force of the air on the balloon. Your Self-Critique: ok Your Self-Critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!