#$&* course phy122 2/3 430Professor Smith,
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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: I think that all of the abbreviations can throw me off. I am going to start writing them all down instead of using the abbreviation. So the above symbols stand for the following: P=pressure difference, A=cross sectional area, L= length, f_net=net force, dW= work done, rho=density, v=velocity. Your Self-Critique Rating: ********************************************* 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: I will need to solve for volume first and then plug it into the formula. 4.8m x 3.8m x 2.8 m=51 m ^3 The density of air is 1.3kg/m^3 I will use the following formula mass=density x volume Which looks like this: 1.3kg/m^3*51m^3=66kg
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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: Your Self-Critique Rating: ********************************************* Question: prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: There is no way I could have figured this out on my own. I do semi-understand now that I have looked at your solution. I know that I am not supposed to look first, but there is no way that I would have known what to do. Density of blood= 1000 kg/m3 The pressure difference is calculated by using this equation: rho g h= 1000kg/ m^3 x 9.80m/s2 x 1.60 m= 15600kg/m/s^2 or 15600 N/m^2 or 15600 Pa Then I will need to convert atm to pascals 1 atm= 1.01325 x 10^5 760 mm of mercury is 133 pascals (101.3kPa/760mm)=133Pa/mm 15600 Pa=15600Pa x (1 mm Hg/133 Pa)=117 mm of Hg 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. STUDENT QUESTION I had to find that conversion online because I know that 1atm is 360mmHg but I could find a conversion for atms to Pascals to make it work from what I knew, so I had to find the 133 online. ??? INSTRUCTOR RESPONSE You should know that 1 atmosphere is about 100 kPa (more accurately 101.3 kPa but you don't need to know it that accurately), and that this is equivalent to 760 mm of mercury. Using these two measures it's easy to convert from one to the other and there's no reason to look for or try to remember a conversion directly beteween mm of mercury and Pa. Specifically the conversion factors are • 101.3 kPa / (760 mm of mercury) = 133 Pa / (mm of mercury) and • (760 mm of mercury) / (101.3 kPa) = .0075 mm of mercury / Pa If you use 100 kPa for the purposes of the problems and tests in this course, as I said before it's OK. If you're ever in a situation outside this class where you need the more accurate figure, it's easy to find. STUDENT QUESTION Do you know if our text tells us this conversion?? INSTRUCTOR RESPONSE The list of equivalent quantities is in Table 10-2 in the 6th edition. This specific conversion isn't given, but the number of Pa and the number of mm of mercury in an atmosphere both are. STUDENT COMMENT The conversion of the units here is very confusing to me. The question didn’t ask for a specific unit set, so I just assumed use the one determined by the answer (kg/ms^2=N). My answer is slightly different, because I actually looked up the density of blood, which is slightly higher than water. INSTRUCTOR RESPONSE kg/m^3 * (m/s^2) * m = kg / (m * s^2), not kg * m/s^2. kg m/s^2 is N. To get kg / (m s^2) you would have to divide kg m/s^2 by m^2. That is, you divide N by m^2, obtaining N / m^2. N / m^2 is the unit of pressure, also called the Pascal. Your Self-Critique: I’ll admit that I had trouble envisioning the formula, but after I looked at your answer I understand the formula better now. Your 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: What I have started doing is looking at the question, figuring out what variables I need to solve for, looking at your solution and then figuring out where you got your numbers. To find the volume of the balloon I will need to use this formula: 4/3pir3 which looks like this: 1.33(3.14)(397.065)=1658m^3 and the mass of the air displaced is about 1.3kg/m^3x1658=2155kg. (where did you get the 1.3kg/m^3) The buoyant force is equal to the magnitude to the force of gravity on the displaced air, or about 2155x9.8m/s^2=21119N. If the total mass of the balloon, including helium is 930 kg then the weight is 930kgx9.8m/s^2=9114N The net force is: 21119N-9114N=12,005N 4.0026 is the atomic mass of helium If the 930 kg doesn’t include the helium we have to account also for the force of gravity on its mass. At about .18kg/m^3 the 1600m^3 of helium will have a mass about 300kg on which gravity exerts an approximate force of 2940 N, so the net force on the balloon would be around 11400N-2940N=8760 (where did you get the 300kg)
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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.. ** 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: I am struggling. I did however make more sense of this question than I have with others. I have a few questions in parenthesis that I did not understand. Your Self-Critique Rating: ********************************************* Question: Openstax: (a) What is the mass of a deep breath of air having a volume of 2.00 L? (b) Discuss the effect taking such a breath has on your body’s volume and density. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I learned that air consists of nitrogen and oxygen so I need to find the atomic masses of each. I need to convert the body’s temperature to K. I will use the following equation to solve for this problem: PV=nRT P is atmospheric pressure, V is 2 L or .002 m^2 and T is the body temperature (around 310 K) N=(100000N/m^2x.002m^3)/(8.31J/mole K) x 310 k= 1/10 mole (where did the 8.31J come from in this equation)
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Given Solution: PV = n R T implies that n = P V / R T. P is atmospheric pressure, V is 2 liters or .002 m^3, and T is body temperature (around 310 K). n = (100 000 N/m^2 * .002 m^3) / ( (8.31 Joules / (mole K)) * 310 k) ). The result is about 1/10 of a mole. Nitrogen makes up most of our air; with its atomic number 7 its atomic mass is 14; it is diatomic so a mole has mass 28 grams. Oxygen makes up most of the rest; with its atomic number 8 and diatomic nature this increases the molar density to around 29 grams. 1/10 of a mole of air therefore has a mass of about 3 grams. Your body's density is close to that of water. Its volume increases by 2 liters with the breath, but its mass increases by only about 3 grams (much less than the 2000 grams that would be gained by an equal amount of water). So your density will decrease. A person with a volume of 100 liters weighs around 220 pounds. Increasing the volume of that person by 2 liters corresponds to an increase of 2% in volume, with negligible increase in mass. This would correspond to about a 2% decrease in density. Your Self-Critique: I have some questions that are listed in parenthesis listed above. I think if you can answer these I will understand the solution better. Your Self-Critique Rating: ********************************************* Question: Openstax: The left side of the heart creates a pressure of 120 mm Hg by exerting a force directly on the blood over an effective area of 15.0 cm^2. What force does it exert to accomplish this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Professor, when I see these questions, it is hard for me to see all the variables I need to solve for. Density of blood: 1000kg/m^3 I will use the following formula to solve for this question: dP=rho g h = 13600kg/m^3 x 9.8 m/s^2 x .120m=15000pa (I understand that the 136000kg/m^3 is the density but how did you come up with that number?)
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Given Solution: A 120 mm column of Hg corresponds to pressure change `dP = rho g h = 13600 kg / m^3 * 9.8 m/s^2 * .120 m = 15000 N/m^2 or 15000 Pa. On an area of 15.0 cm^2 this pressure would exert a force of F = P * A = 15 000 N/m^2 * (15.0 cm^2) = 225 000 N/m^2 * cm^2 = 225 000 N/m^2 * (.01 m)^2 = 22.5 N. Your Self-Critique: I’m not sure that I’m ever going to understand these principles. Your Self-Critique Rating: ********************************************* Question: Openstax (optional for Principles of Physics): What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are parts by mass, not volume.) Assume that this is a simple mixture having an average density equal to the weighted densities of its constituents. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Using very approximate densities (20 000 kg/m^3 for gold, 9000 kg/m^3 for silver and 8000 kg/m^3 for copper) we obtain the following sample solution: 1 part of gold occupies 1/20 the volume of 1 part of water 1 part of silver occupies 1/9 the volume of 1 part of water 1 part of copper occupies 1/8 the volume of 1 part of water. So 18 parts gold, 5 parts silver and 1 part copper will occupy the volume of 18 * 1/20 + 5 * 1/9 + 1 * 1/8 = 569 / 360 = 1.58 (approx.) part of water and will have a mass of 18 + 5 + 1 = 24 times that of 1 part of water. The density will therefore be 24 / 1.58 = 15 times that of water. Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: Optional Openstax: A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I understand that we need to find the area for the cork. I also understand that we need to find fluid pressure from the wind that has been poured into the jug. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The 120-N force exerted on the wine by the cork was spread over an area of about 3 cm^2 (this being the approximate area of a circle 2 cm in diameter), resulting in an additional fluid pressure of 120 N / (3 cm^2) = 40 N / cm^2 = 40 N / (.01 m)^2 = 400 000 N / m^2. As a result the pressure of the wine increased by the same amount. The bottom of the jug has area approximately 150 cm^2 = .015 m^2, so it experienced an additional force of .015 m^2 * 400 000 N/m^2 = 6000 N, well over 1000 pounds. Your Self-Critique: Area of the cork- how do you get that the area of a wine cork would be 3cm^2 when I calculate the area using pir^2 my answer was 7.065cm^2
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Given Solution: ** 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. ** GENERAL STUDENT QUESTION I have completely confused myself on the equations for liquids and gas and cant figure out if they are interchangeable or I have just been leaning the wrong equations for different variables. I mentioned a few throughout the excercise. For example P= F/A = mg/A = rhoAgh/A = rhogh but this is the equation for PE as well? However in some notes PE = rho A g L then other times it = rho g h Is the first equation only used for fluids and the second for gas? "" INSTRUCTOR RESPONSE P = F / A is the definition of pressure (force per unit of area) In a fluid, the fluid pressure at depth h is rho g h. This quantity can also be interpreted as the PE per unit volume of fluid at height h, and is what I call the 'PE term' of Bernoulli's Equation • Bernoulli's Equation that equation applies to a continuous moving fluid, and one version reads 1/2 rho v^2 + rho g h + P = constant.. The 1/2 rho v^2 term is called the 'KE term' of the equation, and represents the KE per unit of volume. This equation represents conservation of energy, in a way that is at least partially illustrated by the exercises below. The equation you quote, PE = rho A g L, could apply to a specific situation with a fluid in a tube with cross-sectional area A and length L, but that would be a specific application of the definition of PE. This is not a generally applicable equation. Self-critique: Your Self-Critique Rating: ********************************************* Question: `q003. The cap is screwed onto jar half-full of water, and the jar is place on a level surface, on its side. The cap of the jar has diameter 8 cm. How much force will water pressure exert on the cap? Note that it is necessary to set up an integral to solve this problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ********************************************* Question: `q003. The cap is screwed onto jar half-full of water, and the jar is place on a level surface, on its side. The cap of the jar has diameter 8 cm. How much force will water pressure exert on the cap? Note that it is necessary to set up an integral to solve this problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: #*&! ********************************************* Question: `q003. The cap is screwed onto jar half-full of water, and the jar is place on a level surface, on its side. The cap of the jar has diameter 8 cm. How much force will water pressure exert on the cap? Note that it is necessary to set up an integral to solve this problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: #*&!#*&!