Recent content by dbakg00

Thanks to all for the help!

When I do superposition, I get the following: (short circuiting the 1V source) V01 = Vs1(-R2/R1) V01 = 2(-10/2) v01 = -4 V (short circuiting the 2V source) V02 = Vs2(-r2/R1) V02 = 1(-10/2.5) V02 = -4 V V0 = V01 + V02 V0 = (-4) + (-4) V0 = -8 how does that look?

So if I let the 1V source = 0, do I still consider both the 5k and the 2.5k resistors? I know that in an inverting amplifier, vo = (-R2/R1)vi. R2 is obviously 10k. What is confusing me is what value I would use for R1. If I did the superposition as you suggested, when I killed the 1V source...

How could they both have the same direction? The negative terminal of the 1v touches the negative terminal of the 2v. Sorry if this is a dumb question...I'm new to circuits.

Homework Statement See attached picture. My objective is to calculate V0 on the 2k resistor. I'm used to seeing op amps that are more simple looking. I'm trying to reduce this op amp circuit to a simpler one (my attempt is the bottom picture). My question is: did I simplify the...

Homework Statement I have attached a pdf of the problem I am working on. The top circuit is the original problem; I am supposed to solve for the voltage drop across the 4-ohm resistor (top right). I HAVE to use the superposition theorem to solve this problem. The 4 circuits drawn below...

I think I may have figured out the answer to question (b) If \delta=\frac{E_{0}-E_{p}}{E_{0}} and E_{o} = Initial Kinetic Energy, which = \frac{1}{2}*m*{v_{0}}^{2} then \delta=\frac{E_{0}}{E_{0}}-\frac{E_{p}}{E_{0}} = 1-\frac{E_{p}}{E_{0}} = 1-\frac{v_{p}}{v_{0}} = 1-V Is this right...

Homework Statement I am working on a simulated lab in which we have a single particle projectile launched at a target particle (located at the center of the circular chamber) of similar weight. Once the collision takes place, I record the time it takes for the scattered projectile to travel...

Thanks for your help.

I see where I made the mistake now, I was making the denominator the entire mass of the system instead of adding the vertical mass to the entire system mass. The acceleration will increase, but not double. Did part b look ok? Thanks

Homework Statement I have a modified atwood machine which is made up of a vertical mass hanger (hanging over the side of a table) connected to a horizontal cart (on the table) via a rope (un-stretchable) and a pulley (massless, frictionless, etc). The cart wheels have negligible friction...

My question about this setup is if you were to double the mass of the hanging weight, without changing the total mass, would the acceleration double also? My thoughts are that the acceleration would not double because the total mass in the system would remain constant. This is more of a...

Ok, now I see, it would take an increase of 4k to double the velocity with a constant radius. I was over thinking it...thanks for you help everyone

k*x=m*r*\omega^{2} and \omega^{2}= \frac{v^{2}}{r^{2}} so i think the force, kx, is equal to \frac{m*v^{2}}{r} If the above is true, then I can see how k is related to the velocity, but I am not quite sure how to get v to double. Maybe multiply k by (1/2)??

by two??