course Phy 122 ˼Řqq~zassignment #003
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18:37:12 query intro set problem 14 substance, water, both temperatures and masses known, final temperature known, find spec ht Explain how the final temperature of the system is combined with the known initial temperatures and masses of both substances to obtain the unknown specific heat
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RESPONSE --> Thermal energy loss of water is determined by taking (J/kg C) multiplying that by the mass and then multiplying that by the difference in the initial and final temperatures. To determine the substance gain per mass unit you take the Joules and divide that by the mass. The specific heat is thus the substance gain per mass unit divided by the lowest temperature(final) and this yields units of J/KG C.
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18:37:24 ** The change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other. For an ideal substance the change in the thermal energy of an object is proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for the change in the thermal energy of an object we can express this as `dQ = mass * specific heat * `dT. (General College and University Physics students note that specific heat is not in fact strictly constant and for most substances changes with temperature.) We know that what energy one substance loses, the other gains, and we know the specific heat of water , so we can set the two quantities equal and solve for specific heat of the unknown substance. For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently m1 c1 `dT1 = - m2 c2 `dT2. If we know all the quantities except, say, the specific heat c2 of the second substance we can easily solve for c2. **
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RESPONSE --> okay
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18:42:13 prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F.
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RESPONSE --> 86 C + 273 = 359 K 78 F - 32 F = 46 F * 5/9 = 26 C -100 C + 273 = 173 K 5500 C + 273 = 2773 K -459 F + -32 = -491 * 5/9 = -273 C + 273 C = 0 K
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18:42:19 The Kelvin temperature is 273 above the Celsius temperature (actually 273.15 below, but the degree of precision in the given temperatures is not sufficient to merit consideration of tenths or hundredths of a degree). 86 C, -100 C and 5500 C are therefore equivalent to ( 86 + 273 ) K = 359 K, -(100 + 273) K = 373 K, (5500 + 273) K = 5773 K. 78 F is (78 F - 32 F) = 46 F above the freezing point of water. 46 Fahrenheit degrees is the same as (5/9 C / F ) * 46 F = 26 C above freezing, or since freezing is at 0 C, just 26 C. The Kelvin temperature is therefore (26 + 273) K = 299 K. -459 F is (459 + 32) F = 491 F below freezing, or (5/9 C / F) * (-491 F) = 273 C below freezing, which is -273 C or (-273 + 273) K = 0 K. This is absolute zero, to the nearest degree.
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RESPONSE --> ok
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18:45:02 prin phy and gen phy Ch 13.30 air at 20 C is compressed to 1/9 of its original volume. Estimate the temperature of the compressed air assuming the pressure reaches 40 atm.
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RESPONSE --> 20 C + 273 = 293 K * 1/9 = 30.3 K
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18:45:35 query gen phy ch 13.38 fraction of air released after tire temp increases from 15 to 38 C at 220 kPa gauge
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RESPONSE --> I really believe the last problem was wrong but there wasn't an explanation after the problem.
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18:45:39 ** T1 = 288 K, T2 = 311 K so T2 / T1 = 311 / 288 = 1.08, approx. This is approx. an 8% increase in temperature. The pressure will end up at 3ll / 288 * 321 kPa = 346 kPa, approx (note that we have to use actual rather than gauge pressure so init pressure is 220 kPa + 101 kPa = 321 kPa, approx. You then have to change the number n of moles of gas to get back to 331 kPa, so n3 / n2 = P3 / P2 = 321 kPa / 346 kPa or approximately .93, which would be about a 7% decrease. Note that the results here are mental estimates, which might not be particularly accurate. Work out the process to see how the accurate numbers work out. **
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18:45:42 query univ phy 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m. How did you calculate the total radiation of the Sun and how did you use this result to get the radiation per unit area?
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18:45:44 ** GOOD STUDENT SOLUTION The total radiation of the sun was the rate it reaches earth times the imaginary surface of the sphere from the sun center to earth atmosphere, or 1500 W/m^2 * (4`pir^2) = 1500W/m^2 * 2.8537 x10^23 m^2 = 4.28055 x 10 ^ 26 W. } Radiation per unit of area surface of the sun would be
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18:45:47 4.28055 x 10 ^ 26 W / (4*`pi * (6.96 x 10 ^ 8 m)^2) = 4.28055 x 10 ^ 26 W / 6.08735 x 10 ^ 18 m^2 = 70318775.82 J/s/m^2 = 7.03 x 10 ^ 7 J/s/m^2 If the sun is radiating as an ideal blackbody, e = 1, the T would be found: H = `dQ/`dt = 4.28055 x 10 ^ 26 W = (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4 So T^ 4 = 4.28055 x 10 ^ 26 W / 6.087351 x 10 ^ 18 m^2) * 1 * (5.67051 x 10^-8 W/m^2*K) T^4 = 1.240 * 10 ^ 15 K ^4 T = 5934.10766 K on surface of sun. **
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18:45:50 univ phy 17.115 time to melt 1.2 cm ice by solar radiation 600 w/m^2, 70% absorption, environment at 0 C.
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18:45:52 ** Thermal energy is not radiating in significant quantities from the ice, so you use 70% of the incoming 600 watts/m^2, which gives you 420 watts / m^2, or 420 Joules/second for every square meter if ice. Melting takes place at 0 C so there is no thermal exchange with the environment. A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can correct this by using the correct density of ice). It takes about 330,000 Joules to melt a kg of ice, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec the time required will be about 10,000 seconds, or around 3 hours. All my calculations are approximate and done mentally so you should check them yourself, using more precise values of the constants, etc. **
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͗ʮ{ސԝz assignment #004 004. `query 3 Physics II 09-13-2008
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18:48:30 query problem 15 introductory problem sets temperature and volume information find final temperature. When temperature and volume remain constant what ratio remains constant?
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RESPONSE --> PV = nRT V / T = nR / P R is a constant so n/P is the constant ratio.
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18:48:35 ** PV = n R T so n R / P = T / V; since T and V remain constant T / V and therefore n R / P remain constant; since R is constant it follows that n / P remains constant. **
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RESPONSE --> ok
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18:50:16 why, in terms of the ideal gas law, is T / V constant when only temperature and volume change?
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RESPONSE --> PV = nRT P / nR = T / V R is constant so the ratio P / n is constant thus leaving a constant T/V as well.
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18:50:33 ** STUDENT ANSWER AND INSTRUCTOR RESPONSE: They are inverselt proportional. They must change together to maintain that proportion. INSTRUCTOR RESPONSE: You haven't justified your answer in terms of the ideal gas law: PV = n R T so V / T = n R / P. If only T and V change, n and P don't change so n R / P is constant. Therefore V / T is constant, and so therefore is T / V. You need to be able to justify any of the less general relationships (Boyle's Law, Charles' Law, etc.) in terms of the general gas law. **
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RESPONSE --> alright
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18:59:07 prin and gen problem 14.3: 2500 Cal per day is how many Joules? At a dime per kilowatt hour, how much would this cost?
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RESPONSE --> 2500 Cal / day * 4200 J / cal = 10,500,000 kw = 1000 J / s kwh = 1000 J / s * 3600 s = 3600000 J 10,500,000 J / 3600000 J/kwh = 3 kwh
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18:59:56 One Cal (with capital C) is about 4200 Joules, so 2500 Cal is about 4200 * 2500 Joules = 13,000,000 Joules. A watt is a Joule per second, a kilowatt is 1000 Joules / second and a kiliowatt-hour is 1000 Joules / second * 3600 seconds = 3,600,000 Joules. 13,000,000 Joules / (3,600,000 Joules / kwh) = 4 kwh, rounded to the nearest whole kwh. This is about 40 cents worth of electricity. It's worth noting that you use 85% of this energy just keeping yourself warm, so the total amount of physical work you can produce in a day is worth less than a dime.
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RESPONSE --> I rechecked my math and I believe mine is correct. I wasn't sure about the cost of a kwh so 1 kw is .10 cents thus for my answers would cost 30 cents of electricity.
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20:38:20 prin phy and gen phy problem 14.07 how many Kcal of thermal energy would be generated in the process of stopping a 1200 kg car from a speec of 100 km/hr?
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RESPONSE --> 100 km/hr * (1000 m / km) * (1 hr/ 3600 s) = 28 m/s KE = .5 (1200 kg)(28 m/s) KE = 16800 J 1 kcal = 4200 J 16800 J / (1 kcal / 4200 J) = 4 kcal
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20:39:19 **STUDENT SOLUTION for 1000 kg car at 100 km/hr WITH ERROR IN CONVERSION OF km/hr TO m/s: The book tells that according to energy conservation initial KE = final KE + heat or (Q) 100km/hr *3600*1/1000 = 360 m/s ** 100 km / hr is 100 * 1000 meters / (3600 sec) = 28 m/s or so. With units your conversion would be 100km/hr *3600 sec / hr *1/ (1000 m / km) = 360 km^2 sec / ( m hr^2). Correct conversion with conversion factors would be 100 km / hr * (1000 m / km) * (1 hr / (3600 sec) = 28 m/s, approx. Otherwise your solution is correct. STUDENT SOLUTION WITH DIFFERENT ERROR IN UNITS Ke=0.5(1000Kg)(100Km)^2 = 5MJ 1Kcal=4186J 5MJ/4186J==1194Kcal INSTRUCTOR COMMENT: Right idea but 100 km is not a velocity and kg * km^2 does not give you Joules. 100 km / hr = 100,000 m / (3600 sec) = 28 m/s, approx. so KE = .5(1000 kg)(28 m/s)^2 = 400,000 J (approx.). or 100 Kcal (approx). **
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RESPONSE --> I'm not seeing the connection where I messed up. The math for KE doesn't match up.
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20:39:24 query gen phy problem 14.13 .4 kg horseshoe into 1.35 L of water in .3 kg iron pot at 20 C, final temp 25 C, spec ht of hossshoe
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20:39:26 ** STUDENT SOLUTION BY EQUATION (using previous version in which the amount of water was 1.6 liters; adjust for the present 1.35 liters): M1*C1*dT1 + M2*C2*dT2 = 0 is the equation that should be used. 0.4kg * 450J/Kg*Celcius *(25celcius -T1) + 0.3kg * 450J/kg*celcius * (25 celcius - 20 celcius) + 1.6kg *4186J/kg*celcius * (25 celcius - 20 celcius) = 0 } Solve for T1, T1 = 214.8 Celsius Solution below is 189.8 C. GOOD STUDENT SOLUTION: This problem is very similar to the first video experiment. You can determine in the same way the thermal energy gained by the water. The pot gains a little thermal energy too,and you have to know how many Joules are gained by a kg of the iron per degree per kg, which is the specific heat of iron (give in text). You can therefore find the energy lost by the horseshoe, from which you can find the energy lost per degree per kg. For this problem I think we are assuming that none of the water is boiled off ass the hot horse shoe is dropped into the water. First I will find the initial temperature of the shoe. Since water's density is 1g/ml, 1 milliliter of water will weigh 1 gram. So 1.35 Liters of water will have a mass of 1.35 kg. 1.35kg of water is heated by 5 degrees The specific heat of water is 4186 J/kg/degrees celsius so 4186 J / kg / C *1.35 kg * 5 C = 28255 J is necessary to heat the water but since it is in equilibrium with the bucket it must be heated too. mass of bucket = 0.30 kg specific heat of iron = 450 J/kg/degrees 450 J / kg / C * 0.30 kg * 5 C = 675 J to heat bucket So it takes 675 J to heat bucket to 25 degrees celsius 28255 J to heat water to 25 degrees celsius so the horse shoe transferred 675+28255 = 28930 J of energy. Mass of horse shoe = 0.40 kg horse shoe is also iron specific heat of iron = 450 J/kg/degree 28930 J / 0.40kg =72,326 J / kg 72,326 J/kg / (450 J / kg / C) = 160.7 C, the initial temperature of the horseshoe. STUDENT ERROR: MISSING COMMON KNOWLEDGE: Estimate 1.60L of water = 1KG. Could not find a conversion for this anywhere. 1 liter = 1000 mL or 1000 cm^3. Density of water is 1 gram/cm^3 so 1 liter contains 1000 g = 1 kg. Alternatively, density of water is 1000 kg / m^3 and 1 liter = .001 m^3, leading to same conclusion. **
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20:39:29 query univ problem 18.60 (16.48 10th edition) 1.5 L flask, stopcock, ethane C2H6 at 300 K, atm pressure. Warm to 380 K, open system, then close and cool. What is the final pressure of the ethane and how many grams remain? Explain the process you used to solve this problem.
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20:39:31 ** use pV = nRT and solve for n n = p V / (r T) = (1.283 *10^5 Pa )(1.50 * 10^-3 m^3 ) / [ (8.36 J / (mol K) )(380 K) ] = .062 mol, approx.. The atomic masses of 2 C and 6 H add up to 30.1, meaning 30.1 grams / mol. So total mass of the gas is initially m(tot) = (.062 mol)(30.1 g/mol) m(tot) = 1.86 g Now if the system is heated to 380 K while open to the atmosphere, pressure will remain constant so volume will be proportional to temperature. Therefore the volume of the gas will increase to V2 = 1.5 liters * 380 K / (300 K) = 1.9 liters. Only 1.5 liters, with mass 1.5 liters / (1.9 liters) * 1.86 grams = 1.47 grams, will stay in the flask. The pressure of the 1.47 grams of ethane is 1 atmosphere when the system is closed, and is at 380 K. As the temperature returns to 300 K volume and quantity of gas will remain constant so pressure will be proportional to temperature. Thus the pressure will drop to P3 = 1 atm * 300 K / (380 K) = .79 atm, approx.. **
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20:39:34 univ phy query problem 18.62 (16.48 10th edition) unif cylinder .9 m high with tight piston depressed by pouring Hg on it. How high when Hg spills over? How high is the piston when mercury spills over the edges?
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20:39:36 ** Let y be the height of the mercury column. Since T and n for the gas in the cylinder remain constant we have P V = constant, and since cross-sectional area remains constant V = A * h, where h is the height of the air column, we have P * h = constant. Thus P1 h1 = P2 h2, with P1 = atmospheric pressure = Patm and h1 = .9 m, P2 = Patm + rho g y. Mercury spills over when the depth of the mercury plus that of the air column is .9 m, at which point h2 = h1 - y. So the equation becomes Patm * h1 = (Patm + rho g y) * (h1 - y). We can solve this equation for y (the equation is quadratic). We obtain two solutions; one y = 0 tells us what when there is no mercury (y = 0) there is no deflection below the .9 m level. The other solution is y = (gh1rho - Pa)/(grho) = .140 m, which tells us that .140 m of mercury will again bring us to .9 m level. We might assume that this level must correspond to the level at which mercury begins spilling over. To completely validate this assumption we need to show that the level of the top of the column will be increasing at this point: The level of the top of the mercury column above the bottom of the cylinder can also be regarded as a function f (y) of the depth of the mercury. If mercury depth is y then the pressure in the cylinder is Patm + rho g y and the height of the gas in the cylinder is Patm / (Patm + rho g y ) * h1. The level of the mercury is therefore f(y) = Patm / (Patm + rho g y) * h1 + y The derivative of this function is f ' ( y ) = 1 - Patmgh1rho/(grhoy + Patm)^2, which is a quadratic function of y. Multiplying both sides by (rho g y + Patm)^2 we solve for y to find that y = sqrt(Patm)(sqrt(g* h1 * rho) - sqrt(Patm) )/ (grho) = .067 m approx., is a critical point of f(y). The second derivative f '' (y) is 2 Patmg^2h1rho^2/(grhoy + Patm)^3, which is positive for y > 0. This tells us that any critical point of f(y) for which y > 0 will be a relative minimum. So for y = .0635 m we have the minimum possible total altitude of the air and mercury columns, and for any y > .0635 m the total altitude is increasing with increasing y. This proves that at y = .140 m the total height of the column is increasing and additional mercury will spill over. To check that y = .140 m results in a total level of .9 m we note that the air column would then be .9 m - .140 m = .760 m, resulting in air pressure .9 / .760 * 101300 Pa = 120,000 Pa. The pressure due to the .140 m mercury column is 19,000 Pa, which when added to the 101,300 Pa of atmospheric pressure gives us 120,000 Pa, accurate to 3 significant figures. The gauge pressure will be 19,000 Pa. A more direct but less rigorous solution: The cylinder is originally at STP. The volume of the air in the tube is inversely proportional to the pressure and the altitude of the air column is proportional to the volume, so the altitude of the air column is inversely proportional to the pressure. If you pour mercury to depth y then the mercury will exert pressure rho g y = 13,600 Kg/m^3 * 9.8 m/s^2 * y = 133,000 N / m^3 * y. Thus the pressure in the tube will thus be atmospheric pressure + mercury pressure = 101,000 N/m^2 + 133,000 N/m^3 * y. As a result the altitude of the air column will be altitude of air column when y cm of mercury are supported: altitude of air column = atmospheric pressure / (atmospheric pressure + mercury pressure) * .9 m =101,000 N/m^2 / (101,000 N/m^2 + 133,000 N/m^3 * y) * .9 m. At the point where mercury spills over the altitude of the air column will be .9 m - y. Thus at this point 101,000 N/m^2 / (101,000 N/m^2 + 133,000 N/m^3 * y) * .9 m = .9 m - y. This equation can be solved for y. The result is y = .14 m, approx. The pressure will be 101,000 N/m^2 + 133,000 N/m^3 * .14 m = 120,000 N/m^2. The gauge pressure will therefore be 120,000 N/m^2 - 101,000 N/m^2 = 19,000 N/m^2. **
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20:39:39 query univ phy 18.75 (16.61 10th edition) univ phy problem 16.61 for what mass is rms vel .1 m/s; if ice how many molecules; if ice sphere what is diameter; is it visible? Give your solution to the problem.
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20:39:42 ** We can solve this problem knowing that ave KE per particle is 3/2 k T so the .5 m v^2 = 3/2 k T, where v is RMS velocity. Thus m = 3 k T / v^2. Then from the density of water and the mass of the particle we can determine its volume, which is equal to 4/3 pi r^3. From this we find r. We obtain volume 3 k T / (v^2 rho), where rho is the density of water. Setting this equal to 4/3 pi r^3 we get r = [ 9 k T / ( 4 v^2 rho) ] ^(1/3). From the mass, Avogadro's number and the mass of a mole of water we determine the number of molecules. The following analysis shows the intermediate quantities we obtain in the process. Some of the calculations, which were done mentally, might be in error so you should redo them using precise values of the constants. At 273 Kelvin we have ave KE = 3/2 k T = 5.5 * 10^-21 Joules. mass is found by solving .5 m v^2 = 3/2 k T for m, obtaining m = 3/2 k T / (.5 v^2) = 5.5 * 10^-21 J / (.5 * (.001 m/s)^2 ) = 1.2 * 10^-14 kg. The volume of the sphere is therefore 1.2 * 10^-14 kg / (1000 kg / m^2) = 1.2 * 10^-17 m^3. Setting this equal to 4/3 pi r^3 we obtain radius r = ( 1.2 * 10^-17 m^3 / 4.2)^(1/3) = ( 2.8 * 10^-18 m^3)^(1/3) = 1.4 * 10^-6 m. Diameter is double this, about 2.8 * 10^-6 m. This is only 3 microns, and is not visible. A water molecule contains 2 hydrogen and 1 oxygen molecule with total molar mass 18 grams = .018 kg. The 1.2 * 10^-14 kg mass of particle therefore consists of 1.2 * 10^-14 / (.018 kg / mole) = 6.7 * 10^-13 moles. With about 6 * 10^23 particles in a mole this consists of 6.7 * 10^-13 moles * 6 * 10^23 particles / mole = 4* 10^11 particles (about 400 billion water molecules). **
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Sْˍ assignment #005 005. `query 4 Physics II 09-13-2008
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21:08:02 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?
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RESPONSE --> force = P*A volume of plug = A * L mass of plug = rho * A * L KE = .5 mass * velocity^2 .5 rho * A * L * v^2 = P*A*L v^2 = P / .5 rho v = `sqrt (P / .5 rho)
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21:08:56 ** 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). **
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RESPONSE --> I didn't multiply by 2 and instead I multiplied by 1/2 P
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21:21:40 prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m.
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RESPONSE --> Mass = density * volume density of air = 1.3 kg/m^3 volume = 4.8 m * 3.8 m * 2.8 m = 51 m^3 mass = 1.3 kg/m^3 * 51 m^3 = 66 kg
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21:21:47 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.
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RESPONSE --> ok
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21:38:49 prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.
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RESPONSE --> density of whole blood = 1.05 * 10^3 pressure = rho * g * h p = 1050 kg/m^3 * 9.8 m/s^2 * 1.60 m p = 16464 kg / (m s^2) p = 16464 kg m / s^2 / m^2 p = 16464 N / m^2 p = 16464 N/m^2 / 133 Pa = 124 mm Hg
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21:42:02 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.
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RESPONSE --> I used the real whole blood density so that allowed for different numbers for the calculations.
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22:10:12 prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force
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RESPONSE --> Vol = 4/3 pi r^3 = 4/3 pi * 7.35^3 = 1665 m^3 1.3 kg / m^3 * 1665 m^3 = 2164 kg 2164 kg * 9.8 m/s^2 = 21210 N 930 kg * 9.8 m/s^2 = 9114 N He mass = 300 kg 300 kg * 9.8 m/s^2 = 2940 N 21210 N - 9114 N - 2940 N = 9156 N m = 9156 N / 9.8 m/s^2 = 934 kg
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22:10:39 ** 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.. **
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RESPONSE --> okay. Same process but different numbers due to rounding
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22:10:42 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.
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22:10:44 ** 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. **
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