Hi,
This is just a force and motion problem with three unknowns, pulled out of Haliday Resnick Walker (7th Edition), Chpt. 6.
17) An initially stationary box of sand is to be pulled across a floor by means of a cable in which the the tension should not exceed 1100 N. The coefficiient...
ok,...
So for the initial theta I end with the following:
2*(sin(inital angle))*(cos(initial angle)) = [(final x)(g)]/(initial velocity)
And with figures:
2*(sin(inital angle))*(cos(initial angle)) = [(45.7)(9.8)]/(460)
Which reduces to...
sin(2*(initial angle)) = [(final...
Hi,
So this is the problem out of Fundamentals of Physics by Haliday Resnick and Walker (Seventh Edition.
By the way this problem would all under a first year calculus based mechanics course.
I know calculus through multivariable...but that's not needed for this...
Hi,
Your given:
m = mass
(fn) = fndamental frequecny
ok now we need the force of tension on the string.
Ok so let's work it out.
(fn) = v/(lambda)
[lambda = wavelength]
And for a string...
(lambda) = 2L
[L = length of string]
thus (lambda) = 2L
and...(fn) =...
Hi,
I actually remember coming across this problem a while back,
if I remember correctly, Itired to solve this but got stuck in trying to apply conservation of energy to the system because I was not able to figure out the [change in height] from initial position to the bottom where...
Hi Jaykawk1,
Ok this kind of problem requires more algebra than physics...
Ok you have to use two principles two solve this
1) (1/2)M(a)(V^2) + (1/2)M(b)(V^2) = (1/2)M(a)(V'(a)^2) + (1/2)M(b)(V'(b)^2)
[ Conservation of total kinetic energy for elastic collisions ]...
Oh yea...and to think I have adobe photoshop on this computer...I could have nice little diagram...well at least now I know I can upload diagrams as attatchments...so that's cool...
BTW would you happen to know a good program for typing out mathematical and physics equations and symbols...
Thanks that makes a lot of sense. :)
(a) I understand,
But would you mind providing an example where you would do (b) if you can think of one...I kind of get it but an example would reassure me that I really understand what you're explaining.
BTW thanks again for the quick reply.
hmn...
Well looking at each step I took everything can be reversed except...the step I took between:
[(x-2) - (x-3)]/x-2 = 2x-3/x-2
||
\/
(x-2)-(x-3)=2x-3
Where I got rid of 'x-2' in the denominator on both sides...so I guess the point you're trying to drive at is that by getting...
Problem: 1 - x-3/x-2 = 2x-3/x-2
OK I know the answer in the back of our textbook is: No Solution
Yet...
I work out the problem to get, x=2
Here is how I got that.
[x-2) - (x-3)]/x-2 = 2x-3/x-2
x-2-x+3=2x-3
4=2x
2=x...
ehh...sorry I couldn't reply earlier.
um when I typed in f(k) I meant f subscript k, which is the coefficient of friction. Not f(k) as in, f*k.
So sorry about that secondly...um I well picked up where I left off and came down to.
.035v^2 + C = dt
But my issue here is what is v? and...