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A rotating pail of water with an angular velocity w has a surface curving up from the centre to the outer edge. Find the shape of the surface. The slope of the surface is dy/dx = tan(theta) at position x is y (= function of x) is the shape of the surface. You must resolve forces to find tan(theta) as a function of x, and then solve for the height, y, as a function of x.
If the pail rotates once per second, and has a diameter of 0.45 m, how high above the centre is the outer edge?
All I see is that w = 2pi rad/sec and r = 0.225 m.
Thanks!
l = r x p for a rotating object
angular momentum = radius times perpendicular momentum
dl/dt = (dr/dt) x p + (dp/dt) x r
The book then says dr/dt equals the instantaneous velocity at that point, but how can the change in distance from the axis of rotation be the same as the change in circular distance?
A man on a small flatcar on a frictionless rail sees 2 trains. A and B approach him from opposite directions, with each train travelling at 60 km/hr. The flatcar, initially at rest, first recoils from a collision with A, then with B, then A again, etc. Assume that the collisions are all perfectly elastic, and that both trains are infinitely more massive than the flatcar.
What is the speed of the flat car after the first collision? the second?
Can some1 explain to me how to do this? bcuz wuldnt an infinitely larger mass equal infinite momentum, therefore wuldnt the flat car go infinite speed after one hit?
Just a quick question. When you have the rate of change of radioactive decay as dN/dt=-(lambda)N.
Then can someone please explain to me why we rearrange this equation to: dN/N=-(lambda)dt.
I don't get how this integrates into N=(No)e^(-(lambda)t). Partially because I havn't done math for the summer, and because I don't get why we put change in N divided by N. Wouldn't change in N divided by N equal the fraction of decrease at any given time? Does this have anything to do with lambda?
also, can some1 plz explain how integrating dv/(g-bv/m) = dt gives u (-m/b)ln((mg-bv)/mg) = t?
Thank you
Prove limx->2 x^3 = 8
let epsilon > 0 be given for x.
|f(x)-L] = |x^3-8|
is < epsilon if o<|x-2|<delta
then |f(x)-L| = |x^3-8| < epsilon
Assuming im right so far what do I do now? Also, can some1 plz explain to me how to relation epsilon to delta, im not sure how and I dont think I did it right on this question.
Suppose that a fraction of mass, 1.00 x10^-12, of terrestrial carbon in living organisms is not carbon 12, but carbon 14, which has a half-life of 5730 years. How many counts per second would you expect from 3.35 kg of pure carbon made from wood taken from a young living Douglas fir tree when u place the material in a well-shielded beta detector?
For some reason I can't use the infinite/square root signs...
1) lim x->1 [x³-(sqrt(x))]/[1-(sqrt(x))]
2) lim x->-infinite (2x-1)/(sqrt(3x²+x+1))
Let f(x) = { x² if x is rational
{ 0 if x is irrational
Prove that lim f(x) = 0
x->0
1) Find the set of points x satisfying:
a) x³ + x² ≤ 2x
b) |3-1/x| < 2
2) Show that any function with domain (-∞,∞) can be written as the sum of an odd function and an even function.
for 1(a) I got it down to: x(x+2)(x-1) ≤ 0. What to do next?
for 1(b): How do u get rid of the absolute value?
A conical tank with radius 6 ft. and height 14 ft that was initially full of water is being drained at a rate of (1/8)√(h). Find a formula for the depth and the amount of water in the tank at any time t.
solve for x:
x^2-5x=logx/log3
Find d²y/dx² by implicit ddifferentiation. x^(1/3)+y^(1/3)=4
(1/3)x^(-2/3)+(1/3)y^(-2/3)d/dx=0
d/dx=((-1/3)x^(-2/3))/((1/3)y^(-2/3))
((1/3)y^(2/3))/((-1/3)x^(2/3))
(y^(2/3))/(-x^(2/3))
(-y/x)^(2/3)
d²y/dx² = (2/3)(-y/x)^(-1/3)((-(d/dx)x+y))/x²
My teacher says the x² in denominator was wrong, or he just didn't get it?
But what do you do after this? I substituted in the y prime and multiplied it out and got:
((-(4/9)x(-y/x)^(-2/3))+(2/3)(-y/x)^(-1/3)y) all over x²
could u simply solving this problem by seeing that 39.2 newtons must be supplied upwards(Fg) and dividing that by mu to get 55 N as the normal force. You can then see than the big box is 6.25 times more mass and therefore requires 6.25 times more force. 6.25 times 55 +55. You get the same answer only it's a hell of a lot easier. I get the other way too tho.
any help on the others?
y=e^(1+lnx) e^(1+lnx) x-¹ (e^(1+lnx))/x The answer in the back is just e. How do you simplify what i got further to e?
y=x^(lnx) i know the derivative of lnx is 1/x but what about x^(lnx)?
ln(cos-¹x) (1/(cos-¹x))(-x/√(1-x²)) -x/(cos-¹x√(1-x²))
The back in the book has the same answer except the -x is a -1 instead.
k i did it again and i got wat u got i forgot to subtract 1 from the exponent(Stupid!), so the teacher is wrong?
I was doin this physics question ,adding vectors and such and got the right answer. What didnt make sense was that the angles of the triangle did not add up to 180 degrees.
I had a triangle with 3 sides (70.7, 40, 33.7) with an angle of 15 degrees between the 70.7 and the 40. The problem is, when I use sine law to figure out the remaining 2 angles, they do not add up to 180 degrees. What is going on here?
tan-¹√(x²-1)+csc-¹x, x>1
d/dx(tan-¹√(x²-1)+d/dx(csc-¹x)
This is where im confused:
I say:
For the first half:
(1/(1+x²-1))(0.5(x²-1)(2x))
(1/x²)(x³-x)
(x³-x)/x²
(x²-1)/x
The second half:
d/dx(csc-¹x)=d/dx(π/2-sec-¹x)
-1/(|x|√(x²-1))
Altogether:
((x²-1)/x)-(1/(|x|√(x²-1))
((x²-1)/x)-(1/(x√(x²-1))
My teacher says instead of (1/(1+x²-1))(0.5(x²-1)(2x)), you get (1/(1+x²-1))2x
One more question:
csc-¹x is just sinx right?
I really have no idea how to do any of these. If some one could show me by example maybe I could catch on?
Find the derivatives:
y=e^-x
y=xe^2-e^x
y=e^sqrtx
y=8^x
y=x^lnx
y=ln(1/x)
y=log(4)x² [the first bracket means base]
y=log(2)(3x+1)
y=log(10)e^x
y=(sinx)^x, 0<x<π/2
the 2a+b=2 is apparently the relationship of a and b according to our teacher. He wants you to make the parabola so that the very tip of it is connected to the x+2
Let f be the function defined as:
f(x) = { x+2 , x < 2
{ ax^2+bx, x >(or equal to) 2
a) what is the relationship between a and b?
I got 2a+b=2
b) find the unique values of a and b that will make f both continuous and differentiable.
i substitued (2-2a) in to b and got down to y=ax^2+2x-2ax, but now what?
A particle moves in a line so its position at any time in t ≥0 is given by the function s(t)=t²-6t+5,
where s is measured in meres and t is measured in seconds.
a) find the displacement during the first 6 seconds.
All i did was input 0 into the equation solve, then do the same with 6 and subtract the two. I got 0. Is this the right way to do that, os is that too "physics like"?
b)Find avg. velocity during first 6 seconds.
i basically did the same thing with (y2-y1)/(x2-x1) and got 0.
c) Find instantaneous velocity when t=4.
the derivative of the function is 2t-6 so:
s(4)=2(4)-6
=2
d) Find the acceleration of the particle when t=4
the derivative of 2t-6 is 2 so is it 2?
e) At what values does the particle change direction?
I got when t=3 but only by looking at the graph. How do you do this by calculus?
f) Where is the particle when s is a minimum?
again by using the graph
4x^3-1.68x10^-5x2+3.36x10^-6-1.68x10^-7=0