A06 - OpenQuery

#$&*

course PHY 232

3/3/10 21:12

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_0

5/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

change in fluid velocity, assuming constant altitude

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Your Solution:

Using bernoulli's law we have (.5 * 'p * v^2) + ('p * g * h)

+ P. Since gravity, density, and altitude are constant the

middle set can be disregarded. We then have (.5 * 'p * v^2)

+ P. The change in each section of the equation is inversely

related to the other. If one increases by X units, the other

will decrease by X units. Therefore we get 'dP = -'d(.5 * 'p

* v^2). Insert the values for the velocity and density and

you can find the change in pressure.

confidence rating #$&*:

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Given Solution:

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.

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). **

Your Self-Critique:

Your Self-Critique Rating:

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Question: 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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Your Solution:

I have not yet completed this assignment but I will take a

stab at the answer.

Is this a top down approach, where X and Y are not affected

by gravity/friction?

If so then no there is not a significant difference.

confidence rating #$&*:

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Given Solution:

** 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. **

Your Self-Critique:

I hadn't done the experiment but X and Y directions seem to

be arbitrary in this case.

Your Self-Critique Rating:

@& The simulation is set up with the balls all moving in the x direction at the beginning, and the purpose is to see how the velocities redistribute after the balls start colliding.*@

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Question: 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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Your Solution:

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Given Solution:

** 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) **

Your Self-Critique:

Your Self-Critique Rating:

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Question: What do you think is the most likely velocity

of the 'red' particle?

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Your Solution:

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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:

Your Self-Critique Rating:

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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:

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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:

Your Self-Critique Rating:

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Question: What do you think the graphs at the right of

the screen might represent?

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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:

Your Self-Critique Rating:

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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:

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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:

Your Self-Critique Rating:

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Question: prin phy and gen phy problem 10.40 gauge

pressure to maintain firehose stream altitude 15 m

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Your Solution:

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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:

Your Self-Critique Rating:

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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:

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:

Your Self-Critique Rating:

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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:

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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:

Your Self-Critique Rating:

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Question: gen phy which term cancels out of Bernoulli's

equation and why?

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Your Solution:

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:

Your Self-Critique Rating:

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Question: univ phy problem 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:

'rho = 1/(1-f)

T = fw

T + dw + V = w

Substitude fw for T

f*dg*V + dw*V = dg*V

divide by V

f*dg + dw = dg

Simplified to

dg = dw/(1-f)

dg/dw = 1/(1-f)

relative density in relation to water

relative density of gold in relation to water is: 19.3

19.3 = 1/(1-f)

solve for f

1-f = 1/19.3

1-1/19.3 = f

f = 0.95

T = fw

T = (0.95)*(12.9) = 12.255

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 needed to look at the solution to finish this one.

Your 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:

As f approaches 1 the limit approaches infinity. This means

that the density gets much larger the more the force

approaches 1N. As f approaches zero the limit approaches 1.

This means that the density becomes 1 as the force gets

smaller.

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:

As f approaches 1 the limit approaches infinity. This means

that the density gets much larger the more the force

approaches 1N. As f approaches zero the limit approaches 1.

This means that the density becomes 1 as the force gets

smaller.

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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&#Your work looks good. See my notes. Let me know if you have any questions. &#