course Phy202 Please ignore the previous file labelled Querry Assinment2 it is included in this file.I wasn't sure on some answers what exactly you were looking for so I gave all the info I worked for the problem. ?{??€???bj???????assignment #002
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20:38:41 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 --> According to the equation for energy conservation, the 'dQ1+'dQ2=0. And since ""dQ is equaltp the specific heat ""c"" multiplied by the mass ""m"" and the change in temperature from T1 to Tf, it can be said that c1*m1*'dT +c2*m2*'dT=0. Where the 'dT is Tf-T1 and Tf-T2 respectively. So, the you can solve for the unknown specific heat c2 for example by c2=-(c1*m1*(Tf-T1))/(m2*(Tf-T2)
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20:40:11 ** 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 --> This is what I staded, but with a little more description.
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20:40:43 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 --> I'm a G-PHY Student
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20:41:21 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 --> Again I'm a General College Physics Student
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20:42:16 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 --> Air at 1atm and 20*C is compressed to 1/9 of its original volume. Compression ratio: 9.0. If the pressure reaches 40atm what is the temperature of the compressed air? (P1*V1)/T1 = (P2*V2)/T2 (9/20)=(40/T2)?2= (40*20/9)=88.89*C
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20:42:38 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 --> A tire is filled with air at 15*C to a gauge pressure of 220kPa. If the tire reaches a temperature of 38*C, what fraction of the original air must be removed if the original pressure of 220kPa is to be maintained? 'dT=(38-15)=23*C (P1*V1)/(T1) = (P2V2)/(T2)?o maintain P1=P2?T2/T1)*V1=V2 Ratio of 'dT = (38/15) or 0.0253% reduction in volume. This will be the same ratio needed in the decrease of volume to maintain the pressure at 220kPa.
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20:46:55 ** 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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RESPONSE --> I made either a common or uncommom mistake by not changing Temp to Kelvin. I see the error of my ways. Thanks
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20:47:23 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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RESPONSE --> G-Phy Stunent
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20:47:39 ** 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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RESPONSE --> G-Student
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20:48:14 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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RESPONSE --> Ok G-Student
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20:48:24 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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RESPONSE --> G-Student
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20:49:03 ** 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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RESPONSE --> G-Student
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w???u?????U??{ assignment #003 ???????????| Physics II 06-20-2006
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21:05:35 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 --> When temperature is held constant PV =constant and when volume is held constant P/T=constant If temperture and volune are held constant P/n=constant
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21:10:04 ** 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 --> I went the opposite direction with the equation PV=nRT I went P/n=RT/V, I see the mistake, I was moving one variable at a time with out considering the new equation
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21:13:00 why, in terms of the ideal gas law, is T / V constant when only temperature and volume change?
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RESPONSE --> In terms of the ideal gas law T / V are constant when only temperature and volume change because n can be held constant with a sealed system, P can be held constant by allowing the volume to change, and R is an universal constant. Also because they are inversively proportional
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21:13:51 ** 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 --> OK I think I did that in my answer
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21:16:20 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 --> 1Cal=4.186kJ and 1=860W a) 2500Cal = 1.05x10^7J b) 2500Cal =2.91kW-h c) 2.91kW-h X .10 = $29.10
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21:20:11 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 worked this problem and rechecked it the back of the book and I guess my self and the quthor missed the point about J/s to 1000J/s to kW-hr
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21:20:59 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 --> How many kcal are generated when the brakes are used to bring a 1200kg car to rest from a speed of 95km/h? ke=(1/2)*(m)(v^2) of hammer mass car=1200kg v of car=95km/h=(95000m/3600s)=26.39m/s Proper method? ke=(1/2)*(1200kg)*(26.39m/s^2)=15.834x10^3J
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21:24:14 **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 --> It was a 1200kg car at 95km/hr I think my answer may be correct
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21:25:10 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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RESPONSE --> A hot iron horseshoe (Mass= 0.40kg), is dropped into 1.35L of water in a 0.30kg iron pot initially at 20*C. If the final equilibrium temperature is 25.0*C, estimate the initial temperature of the hot horseshoe. Q1+Q2+Q3=0?Q1=Q2+Q3 Q1=0.4kg*(450J/kg-C)*(5C-T1) Q2=1.35kg*(4186J/k-C)*(5*C)?.83x10^4J Q3=0.3kg*(450J/k-C)*(5*C)?675J Q1=(2.83x10^4J)+(675J)=28.98x10^3J T1=(28.98x10^3J)/[(0.4kg)*(450J/kg-C)]+25*C=186*C
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21:26:46 ** 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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RESPONSE --> Ok
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21:26:58 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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RESPONSE --> G-Phy
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21:27:04 ** 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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RESPONSE --> OK
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21:27:11 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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RESPONSE --> G-Phy
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21:27:17 ** 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 = (g?1?ho - Pa)/(g?ho) = .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 - Patm??1?ho/(g?ho? + 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) )/ (g?ho) = .067 m approx., is a critical point of f(y). The second derivative f '' (y) is 2 Patm?^2?1?ho^2/(g?ho? + 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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RESPONSE --> OK
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21:27:25 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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RESPONSE --> G-PHY
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21:27:33 ** 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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RESPONSE --> ok
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???y???????…???assignment #004 ???????????| Physics II 06-20-2006
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21:43:58 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 --> The velocity of the exiting water can be determined by taking the difference in pressure multiplied by the cross-sectional area and length of the plug. Multiplying this by 2 and then dividing by the product of density, cross-sectional area and length. Then taking the square-root of the final answer. v = `sqrt( 2 * dP * A * L / (`rho * L * A) ) = `sqrt ( 2 `dP / `rho).
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21:44:41 ** 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 --> That's the sam, a little more simplified, I thought you wanted more detail.
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21:45:11 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 --> G-PHY
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21:45:18 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 --> G-PHY
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21:47:32 prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.
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RESPONSE --> G-PHY
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21:47:37 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 --> G-PHY
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21:56:38 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 --> Wasn't sure what you were asking so I gave everthing. A spherical balloon has a radius of 7.35m and is filled with helium. How large a load can it lift, assuming that the skin and structure of the balloon have a mass of 930kg? Neglect the buoyant force on the cargo volume itself. Balloon r=7.35m V=(4/3)*pi*(r^3)=1.7x10^3m^3 He 'rho=0.179kg/m^3, mass of He=(0.179kg/m^3)*(1.7x10^3m^3)=304.3kg He mg=3x10^3N/m^3 Balloon mg=(930kg)*(9.8m/s)=9.1x10^3N/m^3 Mass of air displaced = (1.29kg/m^3)*(1.7x10^3m^3)=2.2x10^3kg Mass(air)-Mass(He)-Mass(Balloon)=Mass(Load) Mass(load)2.2x10^3kg-304.3kg-930kg=966kg Fb=(m-he+m-balloon+mload)g Fb=(304.3kg+930kg+966kg)9.8m/s=21.45x10^3N/m^2
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21:57:51 ** 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 --> I wasn't sure exactly what you were looking for.
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21:58:00 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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RESPONSE --> G-Phy
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21:58:08 ** 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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RESPONSE --> G-Phy
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gx??????????? assignment #005 ???????????| Physics II 06-20-2006
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22:10:55 query introset change in pressure from velocity change. Explain how to get the change in fluid pressure given the change in fluid velocity, assuming constant altitude
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RESPONSE --> The change in fluid pressure form the change in velocity when altitude is constant can be equated to the fact that `rho g y will not change because there is no change in altitude y. Thereby 0.5 `rho v^2 + P will remain constant. This will make the change in pressure negative of the change in 0.5 `rho v^2.
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22:11:26 ** The equation for this situation is Bernoulli's Equation, which as you note is a modified KE+PE equation. Considering ideal conditions with no losses (rho*gy)+(0.5*rho*v^2)+(P) = 0 g= acceleration due to gravity y=altitude rho=density of fluid }v=velocity P= pressure Constant altitude causes the first term to go to 0 and dissapear. (0.5*rho*v^2)+(P) = constant So here is where we are: Since the altitude h is constant, the two quantities .5 rho v^2 and P are the only things that can change. The sum 1/2 `rho v^2 + P must remain constant. Since fluid velocity v changes, it therefore follows that P must change by a quantity equal and opposite to the change in 1/2 `rho v^2. MORE FORMAL SOLUTION: More formally we could write }1/2 `rho v1^2 + P1 = 1/2 `rho v2^2 + P2 and rearrange to see that the change in pressure, P2 - P1, must be equal to the change 1/2 `rho v2^2 - 1/2 `rho v1^2 in .5 rho v^2: P2 - P1 = 1/2 `rho v2^2 - 1/2 `rho v1^2 = 1/2 rho (v2^2 - v1^2). **
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RESPONSE --> OK
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22:13:09 query billiard experiment Do you think that on the average there is a significant difference between the total KE in the x direction and that in the y direction? Support your answer.
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RESPONSE --> Well that depends on how we relate signigficant, I ran my 30 test and took the average for 10,15,20,25, and 30 and everythime KEy was higher sometimes by a couple of hundred.
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22:14:00 ** In almost every case the average of 30 KE readings in the x and in the y direction differs between the two directions by less than 10% of either KE. This difference is not statistically significant, so we conclude that the total KE is statistically the same in bot directions. **
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RESPONSE --> OK I didn't reference it to precentage I guess I should have.
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22:14:57 What do you think are the average velocities of the 'red' and the 'blue' particles and what do you think it is about the 'blue' particle that makes is so?
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RESPONSE --> I really didn't understand this portion of the experiment.
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22:16:29 ** Student answer with good analogy: I did not actually measure the velocities. the red were much faster. I would assume that the blue particle has much more mass a high velocity impact from the other particles made very little change in the blue particles velocity. Similar to a bycycle running into a Mack Truck. INSTRUCTOR NOTE: : It turns out that average kinetic energies of red and blue particles are equal, but the greater mass of the blue particle implies that it needs less v to get the same KE (which is .5 mv^2) **
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RESPONSE --> Ok that clears up a lot of my understanding. Because I didn't measure the velocities either. My observation was very similar.
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22:16:42 What do you think is the most likely velocity of the 'red' particle?
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RESPONSE --> Again I do not know
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22:17:31 ** If you watch the velocity display you will see that the red particles seem to average somewhere around 4 or 5 **
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RESPONSE --> Oh that's right there were the values at the top of the page. I even had that written in my Lab Book. OOOPPPS
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22:22:21 If the simulation had 100 particles, how long do you think you would have to watch the simulation before a screen with all the particles on the left-hand side of the screen would occur?
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RESPONSE --> The particles didn't seem to collect on the left hand side
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22:22:50 ** STUDENT ANSWER: Considering the random motion at various angles of impact.It would likely be a very rare event. INSTRUCTOR COMMENT This question requires a little fundamental probability but isn't too difficult to understand: If particle position is regarded as random the probability of a particle being on one given side of the screen is 1/2. The probability of 2 particles both being on a given side is 1/2 * 1/2. For 3 particles the probability is 1/2 * 1/2 * 1/2 = 1/8. For 100 particlles the probability is 1 / 2^100, meaning that you would expect to see this phenomenon once in 2^100 screens. If you saw 10 screens per second this would take about 4 * 10^21 years, or just about a trillion times the age of the Earth. In practical terms, then, you just wouldn't expect to see it, ever. **
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RESPONSE --> Ok, you got us
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22:30:02 What do you think the graphs at the right of the screen might represent?
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RESPONSE --> Some form of distribution curve maybe for velocity or KE
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22:30:24 ** One graph is a histogram showing the relative occurrences of different velocities. Highest and lowest velocities are least likely, midrange tending toward the low end most likely. Another shows the same thing but for energies rather than velocities. **
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RESPONSE --> Hey not too bad
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23:00:35 prin phy and gen phy problem 10.36 15 cm radius duct replentishes air in 9.2 m x 5.0 m x 4.5 m room every 16 minutes; how fast is air flowing in the duct?
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RESPONSE --> A-duct='pi(r^2)=7.07m^2 V2=207m^3 t=960s A1v1=V2/t v1=(V2/t)/A1=(207m^3/960s)/7.07m^2=3.04x10^-2m/s
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23:01:36 The volume of the room is 9.2 m * 5.0 m * 4.5 m = 210 m^3. This air is replentished every 16 minutes, or at a rate of 210 m^3 / (16 min * 60 sec/min) = 210 m^3 / (960 sec) = .22 m^3 / second. The cross-sectional area of the duct is pi r^2 = pi * (.15 m)^2 = .071 m^2. The speed of the air flow and the velocity of the air flow are related by rate of volume flow = cross-sectional area * speed of flow, so speed of flow = rate of volume flow / cross-sectional area = .22 m^3 / s / (.071 m^2) = 3.1 m/s, approx.
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RESPONSE --> I think I may have had a decimal point out of place
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23:02:10 prin phy and gen phy problem 10.40 gauge pressure to maintain firehose stream altitude 15 m ......!!!!!!!!...................................
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RESPONSE --> What gauge pressure Pg in the water mains is necessary if a fire hose is to spray water to a height of 15m? P=('rho*g*'dh)= (1.0x10^3kg/m^3)*(9.8m/s)*(15m)= 1.47x10^5N/m^3 Pa= 1.013x10^5N/m^3 Pg = P-Pa Pg= (1.47x10^5N/m^3)(1.013x10^5N/m^3)=4.57x10^4N/m^3
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23:04:00 ** We use Bernoulli's equation. Between the water in the hose before it narrows to the nozzle and the 15m altitude there is a vertical change in position of 15 m. Between the water in the hose before it narrows to the nozzle and the 15 m altitude there is a vertical change in position of 15 m. Assuming the water doesn't move all that fast before the nozzle narrows the flow, and noting that the water at the top of the stream has finally stopped moving for an instant before falling back down, we see that we know the two vertical positions and the velocities (both zero, or very nearly so) at the two points. All that is left is to calculate the pressure difference. The pressure of the water after its exit is simply atmospheric pressure, so it is fairly straightforward to calculate the pressure inside the hose using Bernoulli's equation. Assuming negligible velocity inside the hose we have change in rho g h from inside the hose to 15 m height: `d(rho g h) = 1000 kg/m^3 * 9.8 m/s^2 * 15 m = 147,000 N / m^2, approx. Noting that the velocity term .5 `rho v^2 is zero at both points, the change in pressure is `dP = - `d(rho g h) = -147,000 N/m^2. Since the pressure at the 15 m height is atmospheric, the pressure inside the hose must be 147,000 N/m^2 higher than atmospheric. **
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RESPONSE --> OK
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23:09:44 Gen phy: Assuming that the water in the hose is moving much more slowly than the exiting water, so that the water in the hose is essentially moving at 0 velocity, what quantity is constant between the inside of the hose and the top of the stream? what term therefore cancels out of Bernoulli's equation?
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RESPONSE --> Pressure would be constant and P would cancell out of Bernoulli's equation. .5'rhov^2+'rho(g)y1=rho(g)y2
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23:09:59 ** Velocity is 0 at top and bottom; pressure at top is atmospheric, and if pressure in the hose was the same the water wouldn't experience any net force and would therefore remain in the hose **
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RESPONSE --> OK
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23:10:20 query gen phy problem 10.43 net force on 240m^2 roof from 35 m/s wind. What is the net force on the roof?
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RESPONSE --> If wind blows at 35m/s over a house, what is the net force on the roof if it's area is 240m^2 and is flat? v=35m/s A=240m^2 Air 'rho=1.29kg/m^3 P=(1/2)*('rho)*(v^2)=(0.5)*(1.29kg/m^3)*(35m/s^2)=790.125N/m^3 F=PA?=(790.125N/m^3)*(240m^2)=1.9x10^5N
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23:10:39 ** air with density around 1.29 kg/m^3 moves with one velocity above the roof and essentially of 0 velocity below the roof. Thus there is a difference between the two sides of Bernoulli's equation in the quantity 1/2 `rho v^2. At the density of air `rho g h isn't going to amount to anything significant between the inside and outside of the roof. So the difference in pressure is equal and opposite to the change in 1/2 `rho v^2. On one side v = 0, on the other v = 35 m/s, so the difference in .5 rho v^2 from inside to out is `d(.5 rho v^2) = 0.5(1.29kg/m^3)*(35m/s)^2 - 0 = 790 N/m^2. The difference in the altitude term is, as mentioned above, negligible so the difference in pressure from inside to out is `dP = - `d(.5 rho v^2) = -790 N/m^2. } The associated force is 790 N/m^2 * 240 m^2 = 190,000 N, approx. **
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RESPONSE --> Yes
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23:20:47 gen phy which term cancels out of Bernoulli's equation and why?
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RESPONSE --> 'rho*g*y The hieght doesn't matter
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23:21:02 ** because of the small density of air and the small change in y, `rho g y exhibits practically no change. **
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RESPONSE --> Yes
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23:21:13 univ phy problem 14.67: prove that if weight in water if f w then density of gold is 1 / (1-f). Meaning as f -> 0, 1, infinity. Weight of gold in water if 12.9 N in air. What if nearly all lead and 12.9 N in air?
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RESPONSE --> G-Phy
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23:21:19 ** The tension in the rope supporting the crown in water is T = f w. Tension and buoyant force are equal and opposite to the force of gravity so T + dw * vol = w or f * dg * vol + dw * vol = dg * vol. Dividing through by vol we have f * dg + dw = dg, which we solve for dg to obtain dg = dw / (1 - f). Relative density is density as a proportion of density of water, so relative density is 1 / (1-f). For gold relative density is 19.3 so we have 1 / (1-f) = 19.3, which we solve for f to obtain f = 18.3 / 19.3. The weight of the 12.9 N gold crown in water will thus be T = f w = 18.3 / 19.3 * 12.9 N = 12.2 N. STUDENT SOLUTION: After drawing a free body diagram we can see that these equations are true: Sum of Fy =m*ay , T+B-w=0, T=fw, B=(density of water)(Volume of crown)(gravity). Then fw+(density of water)(Volume of crown)(gravity)-w=0. (1-f)w=(density of water)(Volume of crown)(gravity). Use w==(density of crown)(Volume of crown)(gravity). (1-f)(density of crown)(Volume of crown)(gravity) =(density of water)(Volume of crown)(gravity). Thus, (density of crown)/(density of water)=1/(1-f). **
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RESPONSE --> G-Phy
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23:21:25 univ phy What are the meanings of the limits as f approaches 0 and 1?
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RESPONSE --> G-Phy
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23:21:31 ** GOOD STUDENT ANSWER: f-> 0 gives (density of crown)/(density of water) = 1 and T=0. If the density of the crown equals the density of the water, the crown just floats, fully submerged, and the tension should be zero. When f-> 1, density of crown >> density of water and T=w. If density of crown >> density of water then B is negligible relative to the weight w of the crown and T should equal w. **
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RESPONSE --> G-Phy
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