#$&* course Phy 122 005. `query 5
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Given Solution: ** The net force on the plug is P * A, where A is its cross-sectional area and P the pressure difference. • If L is the length of the plug then the net force F_net = P * A acts thru distance L doing work `dW = F_net * L = P * A * L. If the initial velocity of the plug is 0 and there are no dissipative forces, this is the kinetic energy attained by the plug. The volume of the plug is A * L so its mass is rho * A * L. • Thus we have mass rho * A * L with KE equal to P * A * L. Setting .5 m v^2 = KE we have .5 rho A L v^2 = P A L so that v = sqrt( 2 P / rho). Your Self-Critique: OK Your Self-Critique Rating: oK Question: prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m. Your Solution: V=4.8*3.8*2.8=51m^3 M=d*v so mass=1.3*51=66kg Confidence Rating: 3
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Given Solution: The density of air at common atmospheric temperature and pressure it about 1.3 kg / m^3. The volume of the room is 4.8 m * 3.8 m * 2.8 m = 51 m^3, approximately. The mass of the air in this room is therefore • mass = density * volume = 1.3 kg / m^3 * 51 m^3 = 66 kg, approximately. This is a medium-sized room, and the mass of the air in that room is close to the mass of an average-sized person. Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 1000kg/m^2*9.8m/s^2*1.6m=15680 Pascals/133Pa=117mm mercury confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. 760 mm of mercury is 1 atmosphere, equal to 101.3 kPa, so 1 mm of mercury is 133 Pascals (101.3 kPa / (760 mm) = 133 Pa / mm), so • 15,600 Pa = 15,600 Pa * ( 1 mm of Hg / 133 Pa) = 117 mm of mercury. (alternatively 15 600 Pa * (760 mm of mercury / (101.3 kPa ) ) = 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. The density of blood is actually a bit higher than that of water; if you use a more accurate value for the density of blood you will therefore get a slightly great result. Your Self-Critique: OK Your Self-Critique Rating: OK Question: prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force Your Solution: V=4/3pi r^3=1663 m^3*1.3kg/m^3=2162 kg*9.8m/s^2=21189N 930kg*9.8m/s^2=9114N Fnet=21189N-9114N=12075N 300kg helium*9.8=2940N 12075N-2940N=9135N net force 9135N/9.8=932 kg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The volume of the balloon is about 4/3 pi r^3 = 1660 cubic meters and mass of air displaced is about 1.3 kg / m^3 * 1660 m^3 = 2160 kg. The buoyant force is equal in magnitude to the force of gravity on the displaced air, or about 2160 kg * 9.8 m/s^2 = 20500 Newtons, approx.. If the total mass of the balloon, including helium, is 930 kg then the weight is 930 kg * 9.8 m/s^2 = 9100 N approx., and the net force is • Net force = buoyant force - weight = 20,500 N - 9100 N = 11,400 N, approximately. If the 930 kg doesn't include the helium we have to account also for the force of gravity on its mass. At about .18 kg/m^3 the 1660 m^3 of helium will have mass about 300 kg on which gravity exerts an approximate force of 2900 N, so the net force on the balloon would be around 11,400 N - 2900 N = 8500 N approx. The mass that can be supported by this force is m = F / g = 8500 N / (9.8 m/s^2) = 870 kg, approx.. ** Your Self-Critique: OK Your Self-Critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Question: prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force Your Solution: V=4/3pi r^3=1663 m^3*1.3kg/m^3=2162 kg*9.8m/s^2=21189N 930kg*9.8m/s^2=9114N Fnet=21189N-9114N=12075N 300kg helium*9.8=2940N 12075N-2940N=9135N net force 9135N/9.8=932 kg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 weight is 930 kg * 9.8 m/s^2 = 9100 N approx., and the net force is • Net force = buoyant force - weight = 20,500 N - 9100 N = 11,400 N, approximately. 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.. ** Your Self-Critique: OK Your Self-Critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!