Recent content by Metamorphose

What is Rf --> Op_Amp 1. Consider the amplifier circuit shown. What value of Rf will yield vout = 2V when Is = 10 mA and Ry = 2Rx = 500Ω 2. The way I did this was by employing KCL: (Is that applicable?) The Attempt at a Solution Rx(Is) + Rf(Is) = Vout I'm actually not quite...

Thanks! I was just wondering :)

1. Please refer to attached image. 2. The attempt at a solution. I noted that Vb was grounded and would therefore be equal to 0 and because one of the two op-amp rules states that Va = Vb, both will be equal to 0. I wrote an equation at node Va: [Va - Vin]/R1 + [Va - Vout]/R2 =...

she copied it correctly. I have a copy of the book

I looked at the answer manual, and whether or not it converged was done using limits. The limits shown there were somewhat similar to this. Convergence testing is taught for three more sections, my guess is it has to be shown using limits.

An equation for m2: ƩFy = m2ay ∴m2ay = T - m2g Solving for a in this case gives (T - mag)/m2. If we could find an equivalent for T in terms of m2, we could plug it into the equation for part A. Can this be done by manipulating the same equation to obtain an equation in terms of m2, ay and g?

1. A block with mass m1 is placed on an inclined plane with slope angle α and is connected to a second hinging block with mass m2> m1 by a massless cord passing over a small frictionless pulley. The coefficient of kinectic freiction between mass ma and the incline is negligible. Find: (a) the...

is the answer I provided above correct?

And when this quadratic is solved: t = (2V0M + (4V0m2 + 8da0p)0.5)/2a0P. The negative solution for time t can be discarded. This also alters my answers to b and c

Oh! I forgot to integrate my V0, which should have given me XM = V0Mt + X0M. Which should have been: V0Mt + d = 0.5a0pt2 + v0pt + X0p At which point, solving for t should be solved using a quadratic equation?

I'm not quite sure how to incorporate that into this. I'm sorry

XB = -0.5aBt2 + V0Bt + X0B V0Bt = 0 ∴ XB = -0.5aBt2 + L XB = -0.5aB[(2L)/(aA+ aB]2 + L XB = -aBL/(aA + aB] + L XB = -aBL/(aA + aB) + L(aA + aB)/(aA + aB) XB = aA(L)/(aA + aB)? I was looking over a document I found and in that document, the Bentley was placed at the origin...

1. While driving on the highway at a constant speed v0, significantly above the speed limit, you pass in front of a parked police car without noticing it. After a few seconds, during which time you have moved a distance d, the police car starts chasing you with a constant acceleration a0. Give...

Thank You :) I tend to loose marks due to these silly mistakes. Reworking Part B gives: XB = -0.5aBt2 + V0Bt + X0B V0Bt = 0 ∴ XB = -0.5aBt2 + L XB = -0.5aB[(2L)/(aA+ aB]2 + L XB = -aB/(aA + aB] + L

Because the Bentley will be accelerating to the left, it's acceleration should be labeled as -Ab. Which when plugged in should give: 0.5Aa(t)^2 = -0.5Ab(t)^2 + L Aa(t)^2 + Ab(t)^2 = 2L ---> t^2 (Aa + Ab) = 2L t = [(2L/(Aa + Ab))]^0.5.