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General => The Lobby => Topic started by: strongbad on September 14, 2009, 08:20:50 PM

Title: help me with physics
Post by: strongbad on September 14, 2009, 08:20:50 PM
i dont get this help me

Given the equation F=mv^2/R, what releationship exist between each of the following?
a. F and R
b. F and m
c. V and v

it is probably simple but idk i dont get it

i will probably post more stuff in this thread because this homework assignment is ugh
Title: Re: help me with physics
Post by: The artist formally known on September 14, 2009, 08:25:31 PM
That's like Newton's 2nd law in centripetal acceleration. It's less about physics and more about math.

F = force
m = mass
v = velocity
R = radius

Bold means vector. Where is big V?
Title: Re: help me with physics
Post by: strongbad on September 14, 2009, 08:27:04 PM
Oh for c. I meant F and v my bad. Yeah I understand what the units mean I just dont get how to say what their relationship is.

okay i also dont get

The graph of braking distnace versus car speed is part of a parabola. Thus, the equation  is written d=av^2+bv+c. The distance, d, has units in meters, and velocity, v, has units in meters/second. How could you find the units of a,b, and c? What would they be?

Also an easy question I think. As you can see I suck with variables.
Title: Re: help me with physics
Post by: The artist formally known on September 14, 2009, 08:29:09 PM
That question is worded pretty badly.

a. If F increases, what in the right hand equation must R do? R divides into (m x v^2) to create F. If F goes up, and (m x v^2) stays the same, then R must do what? R must go down or decrease. Do you see that there is an "inverse" relationship between R and F if (m x v^2) remains unchanged?

b. F and m: If F decreases and goes down, and v^2 and R remain unchanged, what must m do? Well, if F decreases, m must also decrease linearly, do you see that?

c. F and v: If F goes up, and m and R remain constant, how will v^2 react? F increases, v must also increase as well. But, because v is squared, it will increase as the "square" to F.

Let's say F = 2 and v = 3: If F increase by 1 to 3, then v increase by 4^2 = 16; if F increases by 1 again, v increases 5^2 or 25. Do you kind of see the "nonlinear" relationship there?
Title: Re: help me with physics
Post by: strongbad on September 14, 2009, 08:31:31 PM
Oh, I get it. Yeah I definitely see a non-linear relationship. Thanks man.
Title: Re: help me with physics
Post by: The artist formally known on September 14, 2009, 08:34:27 PM
Quote from: GUOB on September 14, 2009, 08:27:04 PM
Oh for c. I meant F and v my bad. Yeah I understand what the units mean I just dont get how to say what their relationship is.

okay i also dont get

The graph of braking distnace versus car speed is part of a parabola. Thus, the equation  is written d=av^2+bv+c. The distance, d, has units in meters, and velocity, v, has units in meters/second. How could you find the units of a,b, and c? What would they be?

Also an easy question I think. As you can see I suck with variables.
Here is the start of it, figure out the rest.

(meters) = a(meters/second)2 + b(meters/second) + c
Title: Re: help me with physics
Post by: strongbad on September 14, 2009, 08:41:31 PM
So are a, b, and c, just predeterminded variables that shape the equation? I am familiar with the quadratic equation, so a, b, and c must just be there to shape the parabola?

:(