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course Phy 232

6-8-11 about 11:45 pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

005. `query 5

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Question: 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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Your Solution:

We find the force of the water 1st, which is c-s hole area * pressure. Then find the work done by this force, dW= F*ds ( ds is equal to the length of the plug). Having the amount of work done by the pressure calculated we move on to the mass of the plug. Mass of the plug= c-s hole area * length * density. Using KE= ½ mv^2, solve for v. So, v= sqrt(2*KE/m).

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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.

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Question: prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m.

Your Solution:

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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.

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Question: prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.

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Your Solution:

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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.

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Question: 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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Your Solution:

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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.

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Question: univ 14.57 was 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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Your Solution:

Use the equation P= atmo p + rho*g*dy

P= 1atm + 1000kg/m^3(9.8m/s^2)(.15m)= 1atm + 1470 Pa= 1471 Pa

2nd part, dy= (P-atmo p)/(rho*g)

dy= (1471 Pa - 1atm)/((13.6*10^3 kg/m^3)*9.8m/s^2)= .011m= 1.1cm

so, 15cm - 1.1cm= 13.9cm

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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.

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Question: univ 14.57 was 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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Your Solution:

Use the equation P= atmo p + rho*g*dy

P= 1atm + 1000kg/m^3(9.8m/s^2)(.15m)= 1atm + 1470 Pa= 1471 Pa

2nd part, dy= (P-atmo p)/(rho*g)

dy= (1471 Pa - 1atm)/((13.6*10^3 kg/m^3)*9.8m/s^2)= .011m= 1.1cm

so, 15cm - 1.1cm= 13.9cm

confidence rating #$&*:

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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.

Here's an alternative solution using Bernoulli's equation, somewhat more rigorous and giving a broader context to the solution:

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

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..

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Question: univ 14.57 was 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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Your Solution:

Use the equation P= atmo p + rho*g*dy

P= 1atm + 1000kg/m^3(9.8m/s^2)(.15m)= 1atm + 1470 Pa= 1471 Pa

2nd part, dy= (P-atmo p)/(rho*g)

dy= (1471 Pa - 1atm)/((13.6*10^3 kg/m^3)*9.8m/s^2)= .011m= 1.1cm

so, 15cm - 1.1cm= 13.9cm

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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.

Here's an alternative solution using Bernoulli's equation, somewhat more rigorous and giving a broader context to the solution:

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

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..

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qa51

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