Recent content by neotriz

Never mind, I just calculated the gain for the modulator and demodulator. The reason I thought it would be A/4 is because I remembered the professor gave us an ideal circuit diagram that had a specified gain value for both the modulator and demodulator. Thank you for your input!

Just a quick conceptual question. If you have a sinusoidal signal of 3kHz with an amplitude of A, and it's going through a modulator, and then going through a demodulator, must the output of the modulator resemble exactly to the original signal? The reason why I am asking is because when I...

I forgot that in dot product it gives you scalar result How about this then: I find P to S vector difference , which will result a vector of <-1, -3, -7> and using that vector, I cross product with Q

Homework Statement Given: P=(1,-2,3), Q=(-4,2,5) and S=(2,1-4) Find a unit vector that is perpendicular to triangle PQS Homework Equations Cross and Dot Product The Attempt at a Solution Correct me if I'm doing wrong. I have two solutions that I've thought: 1)What I...

I forgot to mention that this is a DeJong Equation, and I need to implement in the matlab

I just need a quick clarification on how to read this function f(x) = sigma of X^2, starting at j=1, and n so does that mean that f(3) would equal to 36, if n=4?

This is seems like an easy question... Homework Statement A single conservative force acts on a 5.00 kg particle. The equation Fx = (2x + 4) N describes this force, where x is in meters. As the particle moves along the x-axis from x = 2.20 m to x = 6.60 m. Calculate the change in the...

But I thought at x, there is no PE, meaning KE is at max.

So where is the other .5Kx^2 from the otherside is coming from? I'm really sorry, I'm new at this physics thinking

So that means what I wrote before: PE[a]=KE[b]+Work done "The Work Done" is included the Friction's and the Spring's at x?

Here's what I have: PE[a]=KE[b]+Work done by friction (correct)? thus .5K(x)^2 = .5mv^2 + Work done .5(900N/m)(2.00 X 1-^-2 m)^2 = .5(1.6)v^2 + 3.8N(s) NOTE: I changed 2.00 cm to 2.00 X 10^-2 Through alebraic manipulations, I get: .18 = .8v^2 + 3.8(s), where I guess (s) is...

I don't understand When I do the calculation, I have everything except the displacements (the one it is asking me to find) and the max velocity. How would I find the latter one? One more thing: Do we assume at x, PE is zero?

Also, the answer must leave in cm unit. So I guess I have don't have to change the unit when it is compressing 2cm, right?

Thanks :smile: But I still don't get it. If it that was the case, how would I find the velocity at [b]? .

Homework Statement A block of mass 1.6 kg is attached to a horizontal spring that has a force constant 900 N/m. The spring is compressed 2.0 cm and is then released from rest. a) A constant friction force of 3.8 N retards the block's motion from the moment it is released. At what position x...