q is the electric charge of the particle (in coulombs)
and
E is the electric field (in volts per meter)and it equals...capital E as a scalar!
Problem is there is still the question of what the little e is...
Yep, it's advanced physics. And this type of question is not even in the textbook! Another thing that bugs me is the little e. The ONLY mention of the little e in the text is for the rest energy of an electron as being e=.511MeV but looking at the relevant equations I'm given we have mass, time...
Homework Statement
What is the acceleration of a charged particle in a uniform electric field? Assume the particle moves along a straight line parallel to the electric field. Show that a particle starting from rest at x=0 and t=0 the speed and position are given by the following formulas...
After looking for programs for advanced physics with no luck I decided to write my own.
What I have so far is a simple program to calculate relitavistic velocity:
(v1,v2)
Prgm:Local
Input "Enter velocity 1" .v1
Input "Enter velocity 2" .v2
If v1<0 Then
(v1+v2)/(1+(v1*v2))»vr...
It says that given the initial value of a problem:
dy/dx=f(x,y) y(xo)=yo
assume that f and df/dy are continuous fuctions in a rectangle
R={(x,y):a<x<b, c<y<d}
that contains the point (xo,yo). Then the initial value problem has a unique solution #(x)
in some interval...
It says:
In problem 23-28 determine whether Theorem 1 (existence and uniqueness theorem) implies that the given initial value problem has a unique solution.
23. dy/dx=y^4-x^4 y(0)=7
The existence and uniqueness theorem is exactly what this problem is about. As I have stated the book only gives one example and it is nothing like this problem.
I have however looked at the solutions manual for other similar problems and I find in each case the answer book is only concerned...
Thanks Dick. I feel much better that I am not the only one who doesn't understand what is going on. According to the textbook given:
dy/dx=f(x,y) and y(xo)=yo if f and df/dy are continuous then the problem has a unique solution. The initial value of y(0)=7 that is given is supposedly the...
Near as I can tell y'=dy/dx=y^4=x^4 is contiuous for y=0 and y=7
and f(x)=integral(y')=xy^4 -(x^5/5) is continuous for all x so IVP has a unique solution at y(0)=7.
Does this sound right to anyone?
I know that y(0)=7 is a solution to this IVP because I looked in the back of the book. I don't see how to prove it. My textbook only has one example IVP and it is nothing like this.
Homework Statement
I'm having problems with this IVP
dy/dx=y^4-x^4 and y(0)=7 I know the answer is yes but I just don't see how to get it.
Homework Equations
The Attempt at a Solution
I have a problem in my D.E. class that is driving me nuts.
Two drivers, A and B are in a race. Beginning from a standing start they both proceed at a constant acceleration. Driver A covers the last 1/4 of the distance in 3 seconds. Driver B covers the last 1/3 of the distance in 4 seconds...