[SI-LIST] Re: HSDD: Re: SI Position Open READ THIS!!!!
- From: "haq, shan" <haq_shan@xxxxxxx>
- To: "'vadimh@xxxxxxxxxxx'" <vadimh@xxxxxxxxxxx>,"Si-List@Freelists. Org" <si-list@xxxxxxxxxxxxx>
- Date: Thu, 31 Jan 2002 11:38:04 -0500
Guys, please take this off-line!
-----Original Message-----
From: Vadim Heyfitch [mailto:vadimh@xxxxxxxxxxx]
Sent: Thursday, January 31, 2002 11:34 AM
To: Si-List@Freelists. Org
Subject: [SI-LIST] Re: HSDD: Re: SI Position Open READ THIS!!!!
Mary,=20
you apparently misconstrued my message.=20
I actually took the time to corroborate your point with an argument and =
a formula for those, who would like a solid proof. (That's why I did not =
address you personally in my message). It is for their sake as well that =
I emphasized the difference between "velocity" and "speed." Getting =
terminology straight is a good and important start for any technical =
discussion, would not you agree? But, of course, there are other, finer =
aspects of English language that I could still learn about. (or, should =
I properly say instead: "... about which I could still learn"?)
-Vadim
p.s.: ...over and out...end of the thread.
-----Original Message-----
From: Mary [mailto:mary@xxxxxxxxxxxx]
Sent: Thursday, January 31, 2002 11:02 AM
To: 'Si-List@Freelists. Org'
Subject: [SI-LIST] Re: HSDD: Re: SI Position Open READ THIS!!!!
Vadim,
In the English language, it is common to refer to the magnitude of=20
vector quantities by the name of the quantity (e.g. electric field,
acceleration, velocity, etc.). Instead of saying "the magnitude of
the acceleration" or the "magnitude of the force", previous posts
used the words "acceleration" and "force". Why would you take the
time to post a message to the list suggesting that I don't know the
difference between velocity and speed?
Mary
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of Vadim Heyfitch
Sent: Thursday, January 31, 2002 9:37 AM
To: Si-List@Freelists. Org
Subject: [SI-LIST] Re: HSDD: Re: SI Position Open READ THIS!!!!
..Whoops again.....and it never achieves the speed higher than its =3D
initial speed when it separated from the club. Apart from the air drag, =
=3D
the ball's horizontal velocity component remains constant V0x, while its =
=3D
vertical velocity component changes as Vy(t)=3D3DV0y-gt. The speed =
changes =3D
as s(t)=3D3Dsqrt{V0x^2+Vy(t)^2}
The ball finally reaches its original speed - the highest along its =3D
flight trajectory -when it hits the ground.=3D20
If this does not jog your memory, or if the difference between terms =3D
"speed" and "velocity" is not obvious, it's never late to read a =3D
textbook on Mechanics 101 (special edition for EE alumni ;-)
-Vadim
-----Original Message-----
From: Mary [mailto:mary@xxxxxxxxxxxx]
Sent: Thursday, January 31, 2002 10:18 AM
To: 'Si-List@Freelists. Org'
Subject: [SI-LIST] Re: HSDD: Re: SI Position Open READ THIS!!!!
Whoops. Better rethink this. If the acceleration is not positive,
the golf ball will never have a greater velocity than it had=3D20
when it separated from the golf club. Same goes for the car.
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of Kim, Richard
Sent: Wednesday, January 30, 2002 8:05 PM
To: 'arpad.muranyi@xxxxxxxxx'; 'Si-List@Freelists. Org'
Subject: [SI-LIST] Re: HSDD: Re: SI Position Open READ THIS!!!!
I am not sure if you are applying Newton's law correctly.
When you press the clutch, you are removing the rotational
force on the wheels.
Think of hitting a golf ball. After the impact, the applied
force is no longer there. Applied force being mass and=3D20
acceleration of the club head. =3D20
Does the ball reach the highest speed at the impact and
slows down immediately?
Although the applied force has gone away, the ball should continue
accelerating until it meets opposing forces.
Now the force on the ball is the mass x acceleration of the ball.
If there was no air resistance and gravity, according to Newton's law
the ball should have constant acceleration resulting in increase in =3D
speed.
At the point of impact, the ball receives a force =3D3D mass of club =
head
x accel of club head. Due to air resistance and gravitational pull the
the ball immediately starts to decrease its acceleration, but still
increasing
in speed until acceleration becomes zero.
So getting back to car example, if there were no air resistance, =3D
friction
from the wheels and no gravity, the acceleration of the car would be
constant
therefore increasing in speed.
If the acceleration was high enough, it wouldn't be surprising to =
see=3D20
some increase in speed while acceleration goes down to zero.
Richard
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