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course Phy 202
2/25 12Constants:
k = 9*10^9 N m^2 / C^2 qE = 1.6 * 10^-19 C h = 6.63 * 10^-34 J s
energy of n=1 orbital in hydrogen atom: -13.6 eV k ' = 9 * 10^-7 T m / amp atomic mass unit: 1.66 * 10^-27 kg
electron mass: 9.11 * 10^-31 kg speed of light: 3 * 10^8 m/s Avogadro's Number: 6.023 * 10^-23 particles/mole
Gas Constant: R = 8.31 J / (mole K) proton mass: 1.6726 * 10^-27 kg neutron mass: 1.6749 * 10^-27 kg
Problem Number 1
A certain material has density 6.1 kg / liter. If .58 kg of the material are suspended from a string and immersed in a liquid whose density is 1.18 kg / m^3, what will be the tension in the string?
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Density=6.1 kg/L V=M/Density=.58 kg/6.1 kg/L=.095 L
Buoyant force is equal to the weight of the fluid displaced. Tension force is equal to the weight minus buoyant force.
Buoyant force=1000 kg/m^3*.00095 m^3*9.8 m/s^2= 9.3 N
Tension Force=.58 kg *9.8 m/s^2 -9.3= -3.6N
@& Good, but .095 L is .000095 m^3 so the buoyand force is .95 N, not 9.5 N.
Scientific notation would be useful here. 9.5 * 10^-2 liters * .001 m^3 / liter = 9.5 * 10^-5 m^3 so the mass is 9.5 * 10^-5 m^3 * 10^3 kg / m^3 = 9.5 * 10^-2 kg. Multiplied by 9.8 m/s^2 you get the weight to be .93 N.
An alternative in this case would be to use the fact that the mass of a liter of water is 1 kg, so .095 L is .095 kg, with weight .93 N.
In general it's more reliable to write this as the equilibrium condition, where sum of forces = 0. In this case
weight + buoyant force + tension = 0
-.58 kg * 9.8 m/s^2 + .93 N + T = 0*@
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Problem Number 2
Use Bernoulli's equation to determine the pressure change as water flows through a full horizontal pipe from a point where the pipe diameter is .4 meters and velocity 2 m/s to a point where the pipe diameter is .136 meters.
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`rho*g*h +1/2 `rho*v^2+P=constant
Rho=1000 kg/m^3 g= 9.8 m/s^2 v=2 m/s h=0 (stays the same, horizontal)
1000 kg/m^3* 9.8 m/s^2*0 + ˝ (1000 kg/m^3)(2 m/s)^2+`dP=0
2000 kg m/s^2= `dP
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@& You appear to be setting rho g h + 1/2 rho v^2 + `dP = 0.
Bernoulli's equation is
`d (rho g h) + `d(1/2 rho v^2) + `dP = 0,
or equivalently
rho g h + 1/2 rho v^2 + P = constant
or equivalently
rho g h_1 + 1/2 rho v_1^2 + P_1 = rho g h_2 + 1/2 rho v_2^2 + P_2 .
In this situation h is constant, so since v varies P must vary.
Using the given information you find that the velocity on the other side of the constriction is about 20 m/s (this is explained separately, after the solution to Bernoulli's equation). Using Bernoulli's Equation with rho g h = constant, so that `d(rho g h) = 0, we have
`d(1/2 rho v^2) + `dP = 0
so that
`dP = -`d(1/2 rho v^2) =
- (1/2 rho) * (v_2^2 - v_1^2) =
-1/2 rho * ( (20 m/s)^2 - (2 m/s)^2 )
= -1/2 * 1000 kg/m^3 * (-396 m^2 / s^2)
= -396 000 kg / (m s^2)
= -396 000 (kg m / s^2) / m
= -396 000 N / m
= -396 000 Pa.
So get that v_2 = 20 m/s:
You have the information to determine the two values of v. The pipe's cross-sectional area changes by a factor of (.136 m / (4 m) ) ^ 2 = .1, very approximately. The cross-sectional area decreases so the carry the same amount of water the water velocity must increase proportionally. The speed therefore increases from 2 m/s to 2 m/s * (1 / .1) = 20 m/s.
The above is what I'll call the 'proportionality argument'. You can alternatively memorize the continuity equation and apply it:
An equivalent way to calculate this is with the continuity equation
v_1 A_1 = v_2 A_2,
which implies that v_2 = v_1 * (A_1 / A_2).
If r_1 and r_2 are the radii then we have
v_2 = v_1 * (pi r_1^2 / (pi r_2^2) ).
You could go ahead and calculate the actual areas, but the pi divides out and there's no need to involve it:
v_2 = v_1 * (r_1^2 / r_2^2)
You could square the two radiii and complete the calculation in this form, but you could also write this as
v_2 = v_1 * (r_1 / r_2)^2,
which has the advantage that the ratio r_1 / r_2 of radii is the same as the ratio d_1 / d_2 of the diameters (or for that matter the ratio of the circumferences), so that we get
v_2 = v_1 * (d_1 / d_2)^2
and
v_2 = v_1 * (c_1 / c_2)^2
for free.
In this case we can use the diameters so that
v_2 = 2 m/s * (.4 m / (.136 m) ) ^ 2 = 20 m/s,
approximately.
If you understand the proportionality argument continuity equation then it should be easy to understand the continuity equation. However knowing the continuity equation doesn't mean imply understanding of the proportionality argument. I advocate the proportionality argument first, if that argument makes sense to you. If not, then memorizing the continuity equation and using it carefully is an alternative strategy. Ideally you will understand this both ways, and in solving a problem of this nature you will reconcile both points of view and better validate your solution.
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Problem Number 3
A wall is made of a substance whose thermal conductivity is 1.44 J / (m sec Celsius). What is the thickness of the wall if its cross-sectional area is 82 m^2 and if thermal energy flows through the wall at a rate of 45 watts when the inside and outside temperatures are 23 Celsius and -7.001 Celsius?
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Temp Gradient=`dT/`dx=23 C-(-7.001 C)/x
1.44 J/(m sec Cel)*82 m^2 (30.001 C/x)= 45 watts (J/s)
82 m^2(30.001 C/x) = 45 J/s/ 1.44 J (m sec Cel)
30.001 C/x= 31.25 m Cel/82 m^2
30.001 C=.38 Cel/m *x=78.9m
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Problem Number 4
Analyze the pressure vs. volume of a 'bottle engine' consisting of 4 liters of an ideal gas as it operates between minimum temperature 290 Celsius and maximum temperature 480 Celsius, pumping water to half the maximum possible height. Sketch a pressure vs. volume graph from the original state to the maximum-temperature state and use the graph to determine the useful work done by the expansion. Then, assuming a monatomic gas, determine the thermal energy required to perform the work and the resulting practical efficiency of the process.
• University Physics Students: Analyze the entire process, assuming that after the maximum temperature is achieved pressure is suddenly released, resulting in an adiabatic expansion.
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To approach this, would I proceed as we did with the different “states” as in class and solve based on that?
Do I use the KE=3/2nRT equations to solve?
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@& You would find the max height to which water could be raised in the tube, assuming negligible tube volume so that the volume of the gas remains constant. In other words, heating from 290 C to 480 C, starting at atmospheric pressure, how much pressure would be gained and what is the height of a water column that could be supported by this pressure increase?
The system is heated at constant volume to raise water in the tube to half this height, what temperature is required and how high will the water in the tube be?
If water is then allowed to flow from the tube as the gas expands, what will be the change in gas volume, and how much water will therefore be displaced?
What is the PE change of the displaced water?
What is the area beneath the graph?
The gas requires energy change 1/2 n R `dT for every degree of freedom.
If the gas is expanding against constant pressure, it requires another 2 / 2 n R `dT.*@
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Problem Number 5
Analyze the pressure vs. volume of a 'bottle engine' consisting of 4 liters of an ideal gas as it operates between minimum temperature 280 Celsius and maximum temperature 470 Celsius, pumping water to half the maximum possible height. Sketch a pressure vs. volume graph from the original state to the maximum-temperature state and use the graph to determine the useful work done by the expansion. Then, assuming a diatomic gas, determine the thermal energy required to perform the work and the resulting practical efficiency of the process.
• University Physics Students: Analyze the entire process, assuming that after the maximum temperature is achieved pressure is suddenly released, resulting in an adiabatic expansion.
• Repeat for minimum and maximum temperatures 280 Celsius and 660 Celsius. Compare the graphs you obtain and explain why the work done does not increase in proportion to the increased temperature difference.
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Is this the same as the above question?
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@& The procedures are the same.
Note that in this case the gas is diatomic.
This question also asks for a comparison of the graphs and an explanation of the proportionality of the amount of work done.*@
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Problem Number 6
A certain metal has coefficient of linear expansion 20 * 10^-6 / Celsius. By how much would the volume of a container of this metal with a capacity of 250 liters change if the container was heated from 273 C to 333 C? If the ideal gas in the container was allowed to expand into a balloon, would the increased volume of the balloon be significantly affected by the increased capacity of the container?
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Linear—20* 10^-6 Cel
Volume cap—250 liters
Temp change—273 to 333 =60 Cel
dV=(20*10^-6/C)(250L)(60C)=.3 L
Allowed to expand in a balloon—No, the increased volume of the balloon would not be significantly affected by the increased capacity of the container
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@& The coefficient of volume expansion is three times the coefficient of linear expansion, so you would get three times the change in volume, or about .9 liters.
At constant pressure the gas would expand by a certain amount, which you can calculate (ballpark I see about a 10% increase in absolute temperature, which would imply a 10% increase in volume, or about 25 liters). Balloons do operate at some pressure, but the pressure is typically low, so 25 liters is a reasonable ballpark figure.
.9 lters is roughly 4% of this. A 4% difference in volume might be noticeable, and would certainly be measurable, so that answer would depend on what we regard as 'significant'.*@