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Posted: 10/24/2004 8:16:48 PM EST
A car travelling at 50 km/h skids 15 meters. How far does a car skid at 150 km/h?

This is my last bit of 3 hours of homework. I'm sure it's an easy problem, but I stopped caring. First correct answer gets a free kick in the ass.
Link Posted: 10/24/2004 8:17:26 PM EST
42
Link Posted: 10/24/2004 8:24:32 PM EST
[Last Edit: 10/24/2004 8:25:47 PM EST by yugosksfan]
45 meters.

(50kph/15m) = (150kph/x)
cross multiply and solve for x.
Link Posted: 10/24/2004 8:25:09 PM EST
What is the cross sectional area of the car in all aspects? What is the Cd of the car in all aspects? What are the tires Cf at various speeds and angles? Are all 4 tires locked up? All are 4 tires pointed staight ahead? What... is the air-speed velocity of an unladen swallow?

CW
Link Posted: 10/24/2004 8:36:26 PM EST
What is the coefficient of static friction?
Link Posted: 10/24/2004 8:39:41 PM EST
Depends on how much tannerite you used.
Link Posted: 10/24/2004 8:41:42 PM EST
Did you mean to state that the car, initially travelling at 50 km/h skids to a total stop in a distance of 15 m?

For a vehicle traveling at 150 km/h, the aerodynamic drag is a significant force causing the vehicle to slow down when you take your foot off the accelerator. Even the simplest analysis of this will require knowledge of the the coefficient of drag for the car.

My car has anti-lock brakes. It doesn't skid on dry pavement.

European or African swallow? Ni!
Link Posted: 10/24/2004 8:47:27 PM EST
[Last Edit: 10/24/2004 8:47:57 PM EST by brasspile]
Link Posted: 10/24/2004 8:49:57 PM EST
You guys are thinking WAY too hard.

The anwswer is 135.

The kinetic energy of an object is E = 1/2 * m * v^2

The work done sliding something across a surface is E = F*d

Thus the distance slid will be proportional to the square of the velocity.

Call "K" the ratio of velocity squared to distance.

K*50^2 = 15

So K = 15 / 50^2 = 15/2500

Plug in the new velocity and the new distance is K * 150^2 = 15/2500 * 150^2 = 135

Alternatively, scale the original slide distance by the square of the ratios of the velocities

15 * (150/50)^2 = 135


Link Posted: 10/24/2004 8:52:18 PM EST
Too many variables and "what ifs".

Example: The first car MAY have stopped after 15 meters...because it hit a WALL. Logic would lead one to believe that the 150 km/h car would meet the same fate.



-Smartass Nugz
Link Posted: 10/24/2004 8:58:06 PM EST

Originally Posted By Zak-Smith:
You guys are thinking WAY too hard.

The anwswer is 135.

The kinetic energy of an object is E = 1/2 * m * v^2

The work done sliding something across a surface is E = F*d

Thus the distance slid will be proportional to the square of the velocity.

Call "K" the ratio of velocity squared to distance.

K*50^2 = 15

So K = 15 / 50^2 = 15/2500

Plug in the new velocity and the new distance is K * 150^2 = 15/2500 * 150^2 = 135

Alternatively, scale the original slide distance by the square of the ratios of the velocities

15 * (150/50)^2 = 135





OH YEAH ??

Well who's to say that the pavement beyond the 15 meters isnt coated in OIL? Hmm?



This is why I never sought a higher education.

-Nugz
Link Posted: 10/24/2004 9:02:44 PM EST

Originally Posted By Zak-Smith:
You guys are thinking WAY too hard.

The anwswer is 135.

The kinetic energy of an object is E = 1/2 * m * v^2

The work done sliding something across a surface is E = F*d

Thus the distance slid will be proportional to the square of the velocity.

Call "K" the ratio of velocity squared to distance.

K*50^2 = 15

So K = 15 / 50^2 = 15/2500

Plug in the new velocity and the new distance is K * 150^2 = 15/2500 * 150^2 = 135

Alternatively, scale the original slide distance by the square of the ratios of the velocities

15 * (150/50)^2 = 135





Math makes my head explode.

They said I'd did real smart on the ASVAB, 'n then they said all the real smart ones go into the infantry...cuz we're real smart.

Link Posted: 10/24/2004 9:06:51 PM EST
Here is one for you.

You are traveling across 2 consectutive measured miles. For the first of the 2 miles you average 30mph. You then floor it.
How fast do you have to do the second measured mile such that your average speed over both measured miles is 60mph?

ktm500
Link Posted: 10/24/2004 9:11:08 PM EST
[Last Edit: 10/24/2004 9:11:43 PM EST by Mike_Mills]

Originally Posted By Zak-Smith:
You guys are thinking WAY too hard.



On the contrary, Zak-Smith, you are thinking too simplistically. Reality bites. You have been conditioned to provide the type of simplistic answers a high school or undergraduate professor wants. It fails to consider the realities of the situation. Ignoring these realities will lead to inaccuracies and errors.

Consider this:

A bullet rises 3" above the line of sight before striking a target which is 100 m distant. How far above the line of sight must the bullet rise to strike an identical target at a distance of 300 m? Why? Explain.
Link Posted: 10/24/2004 9:12:03 PM EST
[Last Edit: 10/24/2004 9:13:25 PM EST by npd233]

Originally Posted By ktm500:
Here is one for you.

You are traveling across 2 consectutive measured miles. For the first of the 2 miles you average 30mph. You then floor it.
How fast do you have to do the second measured mile such that your average speed over both measured miles is 60mph?

ktm500



Not possible. For your average to work out to 60 mph, you'd have to finish the whole two miles in 2 minutes. It took you 2 minutes to do the first mile.
Link Posted: 10/24/2004 9:13:45 PM EST

Originally Posted By npd233:

Originally Posted By ktm500:
Here is one for you.

You are traveling across 2 consectutive measured miles. For the first of the 2 miles you average 30mph. You then floor it.
How fast do you have to do the second measured mile such that your average speed over both measured miles is 60mph?

ktm500



Not possible.



Depends on the vehicles acceleration, but it would have to be pretty fast.
Link Posted: 10/24/2004 9:27:10 PM EST

Originally Posted By Mike_Mills:

Originally Posted By Zak-Smith:
You guys are thinking WAY too hard.


On the contrary, Zak-Smith, you are thinking too simplistically. Reality bites. You have been conditioned to provide the type of simplistic answers a high school or undergraduate professor wants. It fails to consider the realities of the situation. Ignoring these realities will lead to inaccuracies and errors.


Whatever dude. One uses the context ("domain") of the problem to determine the range of the answer. Given the context in this case is a homework assignment for a basic physics course, that is exactly the kind of answer desired.

A bullet rises 3" above the line of sight before striking a target which is 100 m distant. How far above the line of sight must the bullet rise to strike an identical target at a distance of 300 m? Why? Explain.

If you gave me the sight-over-bore distance and either the BC or the MV, this is merely a boring computational exercise.

-z
Link Posted: 10/24/2004 9:30:11 PM EST

Originally Posted By npd233:

Originally Posted By ktm500:
Here is one for you.

You are traveling across 2 consectutive measured miles. For the first of the 2 miles you average 30mph. You then floor it.
How fast do you have to do the second measured mile such that your average speed over both measured miles is 60mph?

ktm500



Not possible. For your average to work out to 60 mph, you'd have to finish the whole two miles in 2 minutes. It took you 2 minutes to do the first mile.



You got it. They would have to have a time machine like in back to the future.

ktm500
Link Posted: 10/24/2004 9:30:35 PM EST

Originally Posted By Zak-Smith:
You guys are thinking WAY too hard.

The anwswer is 135.

The kinetic energy of an object is E = 1/2 * m * v^2

The work done sliding something across a surface is E = F*d

Thus the distance slid will be proportional to the square of the velocity.

Call "K" the ratio of velocity squared to distance.

K*50^2 = 15

So K = 15 / 50^2 = 15/2500

Plug in the new velocity and the new distance is K * 150^2 = 15/2500 * 150^2 = 135

Alternatively, scale the original slide distance by the square of the ratios of the velocities

15 * (150/50)^2 = 135





what he says.
Link Posted: 10/24/2004 9:36:11 PM EST

Originally Posted By ktm500:
Here is one for you.

You are traveling across 2 consectutive measured miles. For the first of the 2 miles you average 30mph. You then floor it.
How fast do you have to do the second measured mile such that your average speed over both measured miles is 60mph?

ktm500



299,792,458 meters per second would probably be close enough.
Link Posted: 10/24/2004 10:27:37 PM EST
[Last Edit: 10/24/2004 10:34:59 PM EST by brasspile]
Link Posted: 10/24/2004 10:36:35 PM EST

Originally Posted By ktm500:

Originally Posted By npd233:

Originally Posted By ktm500:
Here is one for you.

You are traveling across 2 consectutive measured miles. For the first of the 2 miles you average 30mph. You then floor it.
How fast do you have to do the second measured mile such that your average speed over both measured miles is 60mph?

ktm500



Not possible. For your average to work out to 60 mph, you'd have to finish the whole two miles in 2 minutes. It took you 2 minutes to do the first mile.



You got it. They would have to have a time machine like in back to the future.

ktm500



I'm driving a DeLorean?
Link Posted: 10/24/2004 11:23:08 PM EST
[Last Edit: 10/24/2004 11:24:24 PM EST by TwoStage]
The 150 km skid didn't happen, the crack dealer that was doing 150km while fleeing from the police turned the head lights off to aid on his escape and missed a blind curve hitting the 200 year old oak tree.

Stoping distance was 0 ft., The police never seen the brake lights come on.
Link Posted: 10/25/2004 11:04:28 AM EST

Originally Posted By Zak-Smith:
Whatever dude.
-z



Zak-Smith, now there's an intelligent response, if ever I read one. Of course my original point was that the thought process is important.


Originally Posted By Zak-Smith:

If you gave me the sight-over-bore distance and either the BC or the MV, this is merely a boring computational exercise.

-z



Zak-Smith, of course my original post was about the thought process and the underlying physics. Once you get the physics right there is an important computation needed. If it's boring, well,...hire someone to do it for you.
Link Posted: 10/25/2004 11:16:40 AM EST
The CORRECT answer is a solution to a set of complex partial differential equations that describe the effect of air flow, the friction of the tires on the pavement, etc.
Link Posted: 10/25/2004 11:19:26 AM EST
Exactly, here's another:

A bird is resting on a perch inside a cage. The cage is on a scale. The bird leaves the perch and flies around inside the cage. What happens to the scale's reading. Why? Please explain.
Link Posted: 10/25/2004 11:26:05 AM EST

Originally Posted By ktm500:

Originally Posted By npd233:

Originally Posted By ktm500:
Here is one for you.

You are traveling across 2 consectutive measured miles. For the first of the 2 miles you average 30mph. You then floor it.
How fast do you have to do the second measured mile such that your average speed over both measured miles is 60mph?

ktm500



Not possible. For your average to work out to 60 mph, you'd have to finish the whole two miles in 2 minutes. It took you 2 minutes to do the first mile.



You got it. They would have to have a time machine like in back to the future.

ktm500



So you're saying you'd only have to reach 88mph, then?
Link Posted: 10/25/2004 11:35:32 AM EST

Originally Posted By Zak-Smith:
You guys are thinking WAY too hard.

The anwswer is 135.

The kinetic energy of an object is E = 1/2 * m * v^2

The work done sliding something across a surface is E = F*d

Thus the distance slid will be proportional to the square of the velocity.

Call "K" the ratio of velocity squared to distance.

K*50^2 = 15

So K = 15 / 50^2 = 15/2500

Plug in the new velocity and the new distance is K * 150^2 = 15/2500 * 150^2 = 135

Alternatively, scale the original slide distance by the square of the ratios of the velocities

15 * (150/50)^2 = 135





I use a simplified equation and I also get 135 meters.

skid distance = (Velocity (in kph)^2)/(254(f + G))

f is the coefficient of friction you will solve for using the given data
G is the gradient of the road, which is assumed to be 0.
Link Posted: 10/25/2004 11:45:26 AM EST
[Last Edit: 10/25/2004 11:58:46 AM EST by Zak-Smith]

Originally Posted By Mike_Mills:

Originally Posted By Zak-Smith:
Whatever dude.


Zak-Smith, now there's an intelligent response, if ever I read one.


Commensurate with your own, I think.


Originally Posted By Zak-Smith:
If you gave me the sight-over-bore distance and either the BC or the MV, this is merely a boring computational exercise.

Zak-Smith, of course my original post was about the thought process and the underlying physics. Once you get the physics right there is an important computation needed. If it's boring, well,...hire someone to do it for you.


Using the standard iterative differential method for modelling external trajectory of small arms projectiles through the atmosphere at the primary tool, all someone needs to do to solve your problem is to apply a zero-seeking algorithm to that, but we need the SOB and either the BC or MV to characterize the projectile based on the original trajectory.

-z
Link Posted: 10/25/2004 12:09:08 PM EST

Originally Posted By ArmedAggie:

Originally Posted By Zak-Smith:
You guys are thinking WAY too hard.

The anwswer is 135.

The kinetic energy of an object is E = 1/2 * m * v^2

The work done sliding something across a surface is E = F*d

Thus the distance slid will be proportional to the square of the velocity.

Call "K" the ratio of velocity squared to distance.

K*50^2 = 15

So K = 15 / 50^2 = 15/2500

Plug in the new velocity and the new distance is K * 150^2 = 15/2500 * 150^2 = 135

Alternatively, scale the original slide distance by the square of the ratios of the velocities

15 * (150/50)^2 = 135





I use a simplified equation and I also get 135 meters.

skid distance = (Velocity (in kph)^2)/(254(f + G))

f is the coefficient of friction you will solve for using the given data
G is the gradient of the road, which is assumed to be 0.



As someone who has collected tire skid data for a major tire manufacturer, I can assure you that the f (coefficient of friction) you list above will change for various skid speeds and lengths.

Not enough information is given to solve this problem

Also - your car will generally stop fastest when there is between 10 and 20% skid.
Link Posted: 10/25/2004 12:15:50 PM EST

Originally Posted By Mike_Mills:
Exactly, here's another:

A bird is resting on a perch inside a cage. The cage is on a scale. The bird leaves the perch and flies around inside the cage. What happens to the scale's reading. Why? Please explain.


The scale will read less because the bird is not being measured anymore (it's supported by the air, not the cage).


What do I win?
Link Posted: 10/25/2004 12:19:27 PM EST

Originally Posted By TrollAccount:

Originally Posted By Mike_Mills:
Exactly, here's another:

A bird is resting on a perch inside a cage. The cage is on a scale. The bird leaves the perch and flies around inside the cage. What happens to the scale's reading. Why? Please explain.


The scale will read less because the bird is not being measured anymore (it's supported by the air, not the cage).


What do I win?



Depending on the damper, spring constant, and system mass, the scale will oscillate for a short time. If the scale is very sensitive, the air currents created by the bird flapping around in the cage will cause the scale's reading to fluctuate as well.
Link Posted: 10/25/2004 12:20:32 PM EST
1. The vehicles mass and tire 'footprint' is important.

2. The coefficient of friction of rubber/surface varies with speed.

3. The road surface material and conditions are important.

That's just what I could think of off the top of my head, but I'm sure there's more... types of brakes, application of braking pressure, etc...
Link Posted: 10/25/2004 12:25:20 PM EST
Jesus guys, didnt mean to start a scuffle.

Well Mr. Zak got it. I figured it out this morning in class.

And he's right- it's a basic, retard physics course. Its assumed that everything is taking place in a "perfect physics world". No drag, friction, etc.

Link Posted: 10/25/2004 12:26:43 PM EST

Originally Posted By slacko:
1. The vehicles mass and tire 'footprint' is important.

2. The coefficient of friction of rubber/surface varies with speed.

3. The road surface material and conditions are important.

That's just what I could think of off the top of my head, but I'm sure there's more... types of brakes, application of braking pressure, etc...



What Zak Smith said. I agree that in the real world the coefficient of friction varies somewhat with speed (at high speeds, the rubber melts and the tire "hydroplanes" on the melted rubber). But a common assumption in physics classes is that the coefficient of dynamic friction is constant. (So mention that assumption in your answer).
Link Posted: 10/25/2004 12:31:49 PM EST

Originally Posted By brasspile:
The car problem: 30mph for first mile uses up 2 hoursMINUTES. Writing it down:

Yuo get ∞



Fixed that for ya
Link Posted: 10/25/2004 12:33:53 PM EST

Originally Posted By Zak-Smith:
You guys are thinking WAY too hard.

The anwswer is 135.

The kinetic energy of an object is E = 1/2 * m * v^2

The work done sliding something across a surface is E = F*d

Thus the distance slid will be proportional to the square of the velocity.
...



+1
Link Posted: 10/25/2004 12:37:26 PM EST

Originally Posted By TrollAccount:

Originally Posted By Mike_Mills:
Exactly, here's another:

A bird is resting on a perch inside a cage. The cage is on a scale. The bird leaves the perch and flies around inside the cage. What happens to the scale's reading. Why? Please explain.


The scale will read less because the bird is not being measured anymore (it's supported by the air, not the cage).


What do I win?



What if, instead of a cage, its a sealed container. Will the weight change, then?
Link Posted: 10/25/2004 12:42:11 PM EST

Originally Posted By Zak-Smith:
You guys are thinking WAY too hard.

The anwswer is 135.

The kinetic energy of an object is E = 1/2 * m * v^2

The work done sliding something across a surface is E = F*d

Thus the distance slid will be proportional to the square of the velocity.

Call "K" the ratio of velocity squared to distance.

K*50^2 = 15

So K = 15 / 50^2 = 15/2500

Plug in the new velocity and the new distance is K * 150^2 = 15/2500 * 150^2 = 135

Alternatively, scale the original slide distance by the square of the ratios of the velocities

15 * (150/50)^2 = 135






That only works if energy is conserved. If frictiopn is counted, which it sounds like it is, you have to count the change in energy due to friction, so KEinital== KEfinal + Elost to friction.
Link Posted: 10/25/2004 12:43:01 PM EST

Originally Posted By jkstexas2001:

Originally Posted By TrollAccount:

Originally Posted By Mike_Mills:
Exactly, here's another:

A bird is resting on a perch inside a cage. The cage is on a scale. The bird leaves the perch and flies around inside the cage. What happens to the scale's reading. Why? Please explain.


The scale will read less because the bird is not being measured anymore (it's supported by the air, not the cage).

What do I win?



What if, instead of a cage, its a sealed container. Will the weight change, then?



No. It'll die very soon and won't be able to fly.


Now gimme my prize from the first question damn it!
Link Posted: 10/25/2004 12:45:13 PM EST
[Last Edit: 10/25/2004 12:45:34 PM EST by Zak-Smith]

Originally Posted By notso:

Zak said:
The work done sliding something across a surface is E = F*d


That only works if energy is conserved. If frictiopn is counted, which it sounds like it is, you have to count the change in energy due to friction, so KEinital== KEfinal + Elost to friction.


The assumption in the qustion is that the car comes to rest so KEfinal is zero.
E "lost" to friction is actually the work friction did while it was sliding. F(linear) = u*F(weight downward), where u ("mew") is the friction coeff.

-z
Link Posted: 10/25/2004 3:46:19 PM EST
You're still ignoring the aerodynamic drag. Then what happens if the car starts to slide sideways and the drag coefficient changes (dramatically)?

The answer reflects the depth of thought of the respondent.

All questions should be responded to with, "That depends".
Link Posted: 10/25/2004 4:19:00 PM EST

Originally Posted By Mike_Mills:
All questions should be responded to with, "That depends".


Which is all fine and good until you need to produce an answer with no additional data.
Link Posted: 10/25/2004 4:31:32 PM EST

Originally Posted By Cold_Warrior:
What is the cross sectional area of the car in all aspects? What is the Cd of the car in all aspects? What are the tires Cf at various speeds and angles? Are all 4 tires locked up? All are 4 tires pointed staight ahead? What... is the air-speed velocity of an unladen swallow? hr



African or South American?
Link Posted: 10/25/2004 4:49:18 PM EST
Link Posted: 10/25/2004 5:05:05 PM EST

Originally Posted By TREETOP:

Originally Posted By TrollAccount:

Originally Posted By jkstexas2001:

Originally Posted By TrollAccount:

Originally Posted By Mike_Mills:
Exactly, here's another:

A bird is resting on a perch inside a cage. The cage is on a scale. The bird leaves the perch and flies around inside the cage. What happens to the scale's reading. Why? Please explain.


The scale will read less because the bird is not being measured anymore (it's supported by the air, not the cage).

What do I win?



What if, instead of a cage, its a sealed container. Will the weight change, then?



No. It'll die very soon and won't be able to fly.


Now gimme my prize from the first question damn it!



Your answer was incorect, no prize given. The weight will be the same since the downforce of the bird's wings must equal at least the bird's weight in order for it to fly. The downforce will be transfered to the floor of the cage which is still on the scale.





Yup, just like a helicopter presses downward on the helipad with equal force as it's own weight when it is hovering above it. The problem we now face is that the spread of the weight is related to the height above the landing area, as well as the size of the target. A bird flying around in the belfry of a really tall cage won't affect the weight on the scale nearly as much as in a squatty cage. A helicopter exerts less and less weight on the helipad as it gains altitude above it. At some point the air supporting the flying object does not translate significant force to the surface to measure.
Link Posted: 10/25/2004 5:38:18 PM EST
[Last Edit: 10/25/2004 5:44:21 PM EST by TrollAccount]

Originally Posted By TREETOP:

Originally Posted By TrollAccount:

Originally Posted By jkstexas2001:

Originally Posted By TrollAccount:

Originally Posted By Mike_Mills:
Exactly, here's another:

A bird is resting on a perch inside a cage. The cage is on a scale. The bird leaves the perch and flies around inside the cage. What happens to the scale's reading. Why? Please explain.


The scale will read less because the bird is not being measured anymore (it's supported by the air, not the cage).

What do I win?



What if, instead of a cage, its a sealed container. Will the weight change, then?



No. It'll die very soon and won't be able to fly.

Now gimme my prize from the first question damn it!



Your answer was incorect, no prize given. The weight will be the same since the downforce of the bird's wings must equal at least the bird's weight in order for it to fly. The downforce will be transfered to the floor of the cage which is still on the scale.

Birdshit!

In an open cage MOST of the air forced "downward" by the bird's wing will actually move LATERALLY and POSTERIORLY out of the cage (that's why a bird flys FOREWARD as opposed to straight up) so the weight of the cage WILL lighten when the bird is flying (although it won't be EXACTLY the same as if there were no bird present at all because of some air turbulence). And even a helicopter lifting off the helopad doesn't press down on the pad with EXACTLY the same amount of pressure as at rest because much of that air pressure generated by the prop is deflected laterally by the compressed layer air beneath it. That's how it continues to gain altitude

And as for the second question in a closed container - I also answered it correctly: "No".

Therefore my answer to both questions are correct.

I DEMAND A RECOUNT!!! I'M CALLING MY LAWYERS!!! YOU'RE ALL NOTHING BUT A BUNCH OF CROOKS!!!

GIMME MY DAMN PRIZE!!!


Link Posted: 10/25/2004 7:45:57 PM EST
I never stated whether the cage was open or closed.

Don't forget about turbulence and "frictional" losses in the turbulent airflow.

That depends, is indeed correct and it opens the door to the necessary discussion regarding assumptions one is to make, the size of the parameter space to be investigated in the response,...all sorts of necessary things in order to more completely understand the question, its context and the considerations to be made in developing a technical approach and of course, the answer.

Having not had this discussion, we (collectively) separately approached the problem in many different ways. Only rarely is the one correct answer. Usually only in the academic world.

Link Posted: 10/25/2004 7:50:11 PM EST

Originally Posted By TrollAccount:

Originally Posted By TREETOP:

Originally Posted By TrollAccount:

Originally Posted By jkstexas2001:

Originally Posted By TrollAccount:

Originally Posted By Mike_Mills:
Exactly, here's another:

A bird is resting on a perch inside a cage. The cage is on a scale. The bird leaves the perch and flies around inside the cage. What happens to the scale's reading. Why? Please explain.


The scale will read less because the bird is not being measured anymore (it's supported by the air, not the cage).

What do I win?



What if, instead of a cage, its a sealed container. Will the weight change, then?



No. It'll die very soon and won't be able to fly.

Now gimme my prize from the first question damn it!



Your answer was incorect, no prize given. The weight will be the same since the downforce of the bird's wings must equal at least the bird's weight in order for it to fly. The downforce will be transfered to the floor of the cage which is still on the scale.

Birdshit!

In an open cage MOST of the air forced "downward" by the bird's wing will actually move LATERALLY and POSTERIORLY out of the cage (that's why a bird flys FOREWARD as opposed to straight up) so the weight of the cage WILL lighten when the bird is flying (although it won't be EXACTLY the same as if there were no bird present at all because of some air turbulence). And even a helicopter lifting off the helopad doesn't press down on the pad with EXACTLY the same amount of pressure as at rest because much of that air pressure generated by the prop is deflected laterally by the compressed layer air beneath it. That's how it continues to gain altitude

And as for the second question in a closed container - I also answered it correctly: "No".

Therefore my answer to both questions are correct.

I DEMAND A RECOUNT!!! I'M CALLING MY LAWYERS!!! YOU'RE ALL NOTHING BUT A BUNCH OF CROOKS!!!

GIMME MY DAMN PRIZE!!!





Thats just another way of saying what I said.
Link Posted: 10/25/2004 8:04:30 PM EST
Link Posted: 10/25/2004 8:06:07 PM EST

Prize.

Now.



Or the kitty gets it.


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