Recent content by PascalPanther

Homework Statement I'm trying to determine whether the arrowed nitrogen is acidic or basic. Apparently the answer is that it is acidic. I don't really understand how I am suppose to make that call. That nitrogen is part of secondary amine that is basic, with resonance effect from the oxygen...

(1/2)Mv^2 = (1/2)kA^2 right? v = Sqrt(k/M)A v = Sqrt(710/200kg) * (4.57m) v = 8.6 m/s mv = Mv (150kg)v = (200kg)(8.6 m/s) v= 11.5 m/s mgh = (1/2)mv^2 (150kg)(9.8)h = (1/2)(150kg)(11.5m/s)^2 h = 6.75 m?

Hmm... Before the collision, the block would be: mgh = (1/2)mv^2 Then mv = Mv_2 Is there a way to use this relationship without velocity data (actual height)?

Homework Statement Homework Equations E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2 omega = Sqrt(k/m) f = omega/(2pi) = 1/(2pi)*Sqrt(k/m) The Attempt at a Solution m = 150kg M = 150kg + 50kg = 200kg x = 15ft = 4.57m h1 = 3.04 m h_real = ? x = A? mgh = (1/2)Mv^2 + (1/2)kx^2 Since there is...

Here is what I did initially: v = 3.77m/1.6s mgh = (1/2)mv^2 + (1/2)*I*omega^2 omega = v/R (20kg)(9.8)(3.77m) = (1/2)(20kg)(2.36 m/s)^2 + (1/2)I((2.36)^2/(0.60)^2) I = 88.3 ... which is more than I = MR^2 = 25.2 ... So this is entirely wrong, because I know it should be less, since the...

"Consider the bicycle wheel is not turning initially. A block of mass m is attached (with a massless string) to a wheel. The block is allowed to fall a distance of h. Assume that the wheel has a moment of inertia I about its rotation axis. Find the angular speed of the wheel after the block...

Well, it looks like there is an alternative version of this problem with everything the same except it asks you to find out the rate of rotation and the mass of the generator. So there must be a way somehow to find the mass. Or this problem is just really weird. :bugeye: But it would be...

This question really has me stumped. Am I right to just ignore the fact that there is an inner control room ring? This is what I think I will need: We haven't gotten to inertia yet, but I think this is an inertia problem. I = MR^2 (if I take out the inner ring and say it is a thin-walled...

So when I'm dividing by the volume, this isn't a normal math operation where I have to apply it to all terms right? It only applies to the mass? I am kind of just substituting it looks like. p(u)*a = (p(u)-p(w))*g So I guess if that was the case, the answer is real near the same density...

I'm still trying to get my head around this. So it sounds like: F = ma = F (mass of unknown)*(acceleration) = ((mass of unknown)-(mass of displaced water))*g a = 0.0096 m/s^2 (1/2at^2 = .12 m) (density of unknown) = (mass unknown)/(volume of unknown) (density of water) = (mass of...

I am suppose to find the density of a thick syrup-like liquid (does not disperse). The only information I have is that when a small portion is poured into a cup of unknown volume, it takes 5 seconds for it to reach the bottom. The cup has a height of 12 cm. I am not sure how to approach...

I thought I understood this problem, but my answer is wrong. "In a sample of 18-karat gold, 75 percent of the total mass is pure gold, while the rest is typically 16 percent silver and 9 percent copper. If the density of pure gold is 19.3 g/cm^3, while the densities of silver and copper are...

Since my first way produces a huge equation, how would I go about moving the L to one side for the second equation? I think I must be missing some algebra trick. I cannot find a way to get (L-x) and L to work.

(1/2)*m*v^2 + W(other) = m*g*h (1/2)*m*v^2 + mu*(m*g*cos(theta))*L = m*g*L*sin(theta) ((1/2)*v^2)/L = -mu*(g*cos(theta)) + g*sin(theta) L/((1/2)*v^2) = 1/(-mu*(g*cos(theta))) + 1/(g*sin(theta)) L = ((1/2)*v^2)/(-mu*(g*cos(theta))) + ((1/2)*v^2)/(g*sin(theta)) Plugging in how I got...

h is equal to (L+x(c))*sin(theta)? Or is there something else? I don't think I could solve for L if that was the case...