Recent content by daivinhtran

It's a very interesting problem. My teacher assumes that the moment of inertia of the child is MR^2 But it turn out the translational K = rotational K ==> total K = mv^2 And some other folks (big guys) said that my teacher is wrong " What contributes to the total kinetic energy of the system...

this way doesn't work because different points of the swing have different velocity. They only have same rotational velocity.

Picture is attached

This is not a homework question. An adult exerts a horizontal force on a swing that is suspended by a rope of length L, holding it at an angle θ with the vertical. The child in the swing has a weight W and dimensions that are negligible compared to L. The weights of the rope and of the seat...

we're probably not supposed to ignore the x in this one, right?

Yes, I did, sir. My teacher also checked it but she had no clue why it's like that.. SHe also did the same calculation and get the same pH.

For V of NaOH=0mL The equilibrium equation is HF <=> H+ + F- .1M -x +x +x .1 x x (x^2)/.1 = Ka = 6.8e-4 x= .008246 = [H+] pH = 2.083745544 With V of Naoh=1mL HF + OH- <=> F- + H20 (.02)(.1)...

Buffer (pH of a weak acid decreases when we add NaOH! Yes, really.. This is not homework, but just something that I find not following general rule. Just wondering why. So We have the titration of 20.0 mL of .100M HF with .100M NaOH With 0mLof NaOH, the pH=2.08 (by calculating) But...

I did try this way...I find the region at the interval [-1, 0] by take the integrate of pi ∫(-x+3)^2 - (x^2 - 3x)^2 dx... Then, pi x ∫ (-x+3)^2 - (-x^2 + 3x)^2 dx for interval [0,1] Then, pi x ∫ (3x-x^2)^2 for the interval [1,3] and take the sum of all...and get 56pi/3 but the answer is...

what do you do with the overlapping region?

oh yes...I actually left something out...the region revolve about a, x-axis, and b, y=axis..

Homework Statement THe region bounded by y = -x + 3 and y = x^2 - 3x the region revolve about a, x-axis, and b, y=axisHomework Equations V = π∫r^2 dxThe Attempt at a Solution I have no clue to solve it since the volume overlap. I try to ignore the overlapped region but didn't get the right...

A. (0.98-1)(a/3) + aln(2.98) B. (a/3)aln(2.98) + .98 C. -.02a/3 D. (.98)ln(a/3) E. (a/3) (.98) + aln(2.98) none of them has aln3 like I solved

Homework Statement consider the function f(x) = aln(x+2). Given that f'(1) = a/3, what is the approximate value of f(0.98)?Homework Equations f(x1) = f(x0) + f'(x0)x(x1-x0)The Attempt at a Solution I solved it and get f(.98) = aln(1+2) + (.098-1) = aln(3) - (.02)(a/3) <= not an answer the...

So what happen if fs > μsFn?