Assingment 06

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course PHY 232

f 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.

006. `query 5

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Question: query introset change in pressure from velocity change.

Explain how to get the change in fluid pressure given the initial and final fluid velocities, assuming constant altitude

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Your Solution:

For this situation will turn back to bernoullis eqn as shown below;

1/2*rho*v1^2+rho*g*y1+P1=1/2*rho*v2^2+rho*g*y2+P2

And from the question they tell us that the altitude is constant so from the eqn above we can cancel y1 and y2 leaving us with

1/2*rho*v1^2+P1=1/2*rho*v2^2+P2

We can rearrange the eqn by subtracting P1 from both sides

1/2*rho*v1^2-1/2*rho*v2^2=P2-P1

Therefore, dP=P2-P1=1/2*rho(v1^2-V2^2).

confidence rating #$&*:

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Given Solution:

Bernoulli's Equation can be written

1/2 rho v1^2 + rho g y1 + P1 = 1/2 rho v2^2 + rho g y2 + P2

If altitude is constant then y1 = y2, so that rho g y1 is the same as rho g y2. Subtracting this quantity from both sides, these terms disappear, leaving us

1/2 rho v1^2 + P1 = 1/2 rho v2^2 + P2.

The difference in pressure is P2 - P1. If we subtract P1 from both sides, as well as 1/2 rho v2^2, we get

1/2 rho v1^2 - 1/2 rho v2^2 = P2 - P1.

Thus

change in pressure = P2 - P1 = 1/2 rho ( v1^2 - v2^2 ).

Caution: (v1^2 - v2^2) is not the same as (v1 - v2)^2. Convince yourself of that by just picking two unequal and nonzero numbers for v1 and v2, and evaluating both sides.

ALTERNATIVE FORMULATION

Assuming constant rho, Bernoulli's Equation can be written

1/2 rho `d(v^2) + rho g `dy + `dP = 0.

If altitude is constant, then `dy = 0 so that

1/2 rho `d(v^2) + `dP = 0

so that

`dP = - 1/2 rho `d(v^2).

Caution: `d(v^2) means change in v^2, not the square of the change in v. So `d(v^2) = v2^2 - v1^2, not (v2 - v1)^2.

STUDENT SOLUTION: 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 disappear.

(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.

@&
As it turns out the average KE of particle, as aveaged over a large number of collisions, doesn't depend on its mass. So what you did was OK.
*@

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Question: What do you think is the most likely velocity of the 'red' particle?

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Your Solution:

Using the data from the screen i took the KE x and used it to calculate an approximated

value of the red velocity and i obtained the reasult below

KE x=1451

KE=.5*m*v^2=0.5(64)v^2=1451

= 6.73 and since there are units given i did not include any units.

But by observation i obtained a range between 4 0r 6

confidence rating #$&*:

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Given Solution:

** If you watch the velocity display you will see that the red particles seem to average somewhere around 4 or 5 **

Your Self-Critique:

I think the given solution is right by observation although it would be interesting to find out if this numbers can be backed up by any form of calculations.

Your Self-Critique Rating: Ok

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Question: 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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Your Solution:

Due to the random movements of the particles, we can use probability, we can say 0 to 1. Meaning it might not happen.

confidence rating #$&*:

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Given Solution:

** 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. **

Your Self-Critique:

The given solution is well thought and backed up as well.

Your Self-Critique Rating: Ok

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Question: What do you think the graphs at the right of the screen might represent?

The graphs on the right might show velocity range and the other will show energy.

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Your Solution:

confidence rating #$&*:

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Given Solution:

** 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. **

Your Self-Critique:

The given solution has the same idea.

Your Self-Critique Rating:Ok

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Question: 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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Your Solution:

We start by finding the volume = (9.2m)(5m)(4.5m)=207m^3

convert the min to sec (16min)(60sec)=960sec

Air is replentished at 207m^3/960sec=0.22m^3/sec

Area=(pi)(r^2)=(pi)(0.15m^2)=0.071m^2

We know V=A*v therefore v=V/A=0.22m^3/sec/0.071m^2=3.1m/s

confidence rating #$&*:

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Given Solution:

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.

Your Self-Critique:

The given solution has a similar idea as to how approach the problem.

Your Self-Critique Rating: ok

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Question: prin phy and gen phy problem 10.40 What gauge pressure is necessary to maintain a firehose stream at an altitude of 15 m?

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Your Solution:

We know P=(rho)(g)(h)=(1000kg/m^3)(9.80m/s^2)(15m)=147000 N/m^2

confidence rating #$&*:

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Given Solution:

** 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. **

Your Self-Critique:

The given solution is the same as mine.

Your Self-Critique Rating:Ok

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Question: 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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Your Solution:

since there is 0 velocity then there will be 0 at the top or bottom

confidence rating #$&*:

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Given Solution:

** 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 **

Your Self-Critique:

Similar approach to the problem and the reasult.

Your Self-Critique Rating:Ok

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Question: 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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Your Solution:

For this problem we return to bernoullis equation (0.5)(rho)(v1^2)=(0.5)(rho)(v2^2) and since we are given v2=35m/s then v1=0m/s. Since the Air mass density is

1.29 kg/m^3 then we can substitute these figures into the second part of the equation since the other one cancels out.Therefore, (0.5)(1.29kg/m^3)(35m/s^2)=790N/m^2

Fnet=(790N/m^2)(240m^2)=190000N

confidence rating #$&*:

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Given Solution:

** 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. **

Your Self-Critique:

The given solution has a different approach but similar idea as to how to solve the problem.

Your Self-Critique Rating:OK

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Question: gen phy which term 'cancels out' of Bernoulli's equation and why?

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Your Solution:'

we know that air has a small density so there isnt much of a change in altitude.

confidence rating #$&*:

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Given Solution:

** because of the small density of air and the small change in y, `rho g y exhibits practically no change. **

Your Self-Critique:

Given solution is well presented.

Your Self-Critique Rating:ok

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Question: univ phy problem 12.77 / 14.75 (11th edition 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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Your Solution:

We can start by Buoyant force and tension on rope that is supporting the crown in water

T=fw

f * dg * vol + dw * vol = dg * vol

If we further divide by vol and rearrange the equation the we end up with

f * dg + dw = dg

Hence we can solve for dg=dw/(1-f)

But gold density is 19320kg/m^3 and solving for f= 18320/19320kg/m^3=0.95

Weight=0.95*12.9=12.3N

confidence rating #$&*:

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Given Solution:

** 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). **

Your Self-Critique:

I think the given solution is well presented and well detailed.

Your Self-Critique Rating:ok

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Question: univ phy What are the meanings of the limits as f approaches 0 and 1?

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Your Solution:

For f to approach 0 then the density of crown/density of water must equal 1

confidence rating #$&*:

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Given Solution:

** 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. **

Self-critique (if necessary):

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Self-critique rating:

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Question: univ phy What are the meanings of the limits as f approaches 0 and 1?

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Your Solution:

For f to approach 0 then the density of crown/density of water must equal 1

confidence rating #$&*:

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Given Solution:

Self-critique (if necessary):

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Self-critique rating:

#*&!

Assingment 06

@& As it turns out the average KE of particle, as aveaged over a large number of collisions, doesn't depend on its mass. So what you did was OK. *@

&#Very good responses. Let me know if you have questions. &#

@&
As it turns out the average KE of particle, as aveaged over a large number of collisions, doesn't depend on its mass. So what you did was OK.
*@