Recent content by photonsquared

AlphaQ try quiver3 in Matlab: A three-dimensional quiver plot displays vectors with components (u,v,w) at the points (x,y,z). quiver3(x,y,z,u,v,w) plots vectors with components (u,v,w) at the points (x,y,z). The matrices x,y,z,u,v,w must all be the same size and contain the corresponding...

I agree with Russ. For your mechanical explosions (no chemical reaction) the rupture pressure of the cans is the pressure you need to calculate, which can be approximated using hooks law and the material properties of the container. The firework is a different matter. In general you can...

Engineering Circuit Analysis (Hayt, Kemmerly, Durbin) I found the https://www.amazon.com/dp/007286611X/?tag=pfamazon01-20 above to be very useful. I have used it in two online EE classes (no instructor) and found I could follow it easily.

Thanks for not giving up on my question. Kevin

Well my answer is different from that given in the book. My answer is too high by 1V.

Find v(t) at t=800ms for the circuit in Figure 1. Ans: 802mV Writing a single node equation we have \frac{v(t)-2tu(t)}{5}+0.1\frac{dv}{dt}=0. Taking the Laplace transform we have L\left\{ \frac{v(t)-2tu(t)}{5}+0.1\frac{dv}{dt}=0\right\} . tu(t)\Rightarrow\frac{1}{s^{2}}...

Thanks, I'll attempt again.

1. Find v(t) if V(s)=\frac{2s}{(s^{2}+4)^{2}} Ans: v(t)=\frac{1}{2}tsin2tu(t) 2. Homework Equations : V(s)=\frac{a_{n}}{(s-p)^{n}}+\frac{a_{n-1}}{(s-p)^{n-1}}+\cdots+\frac{a_{1}}{(s-p)} a_{n-k}=\frac{1}{k!}\frac{d^{k}}{ds^{k}}[(s-p)^{n}V(s)]_{s=p} 3. Attempt at a solution...

Perhaps you are answering your own question, in that you can determine the height above the flat plane the block would rising using mgh. Without any additional information regarding the curve I think that answer would suffice.

The work is due to the Force of gravity plus the Force of friction. Not sure what your third force is in your equation. Your equation is correct for the force due to gravity.

Work Assuming the coordinate system is established with the x-axis parallel with the ramp you would use the following: W = F*d where F = Fgravityx + Ff where Fgravityx = mgsin(25) Ff = u*N and u = coefficient of friction N = mgcos(25) W = d(Fgravityx + Ff) Hope this helps

That is the same value I calculate. The reason you don't need the mass of the puck is because the masses cancel out when you substitute eqn 7 into 4. Hope this helps.

I have attached a re-draw of the circuit. Hope this helps. I solved it using three KVL loop equations one for each loop and ended up with three equations and three unknowns. I got 3A to the left or down in the redrawn circuit.

Try this Force of friction is given by: F_{f} = \mu N [1] Solving for \mu we have \mu = \frac{F_{f}}{N} [2] The normal force N is given by: N = mg [3] Substituting [3] into [2] we have: \mu = \frac{F_{f}}{mg} [4] The force of friction is also defined by: F_{f} = ma...

In the standard equation k would be be y intercept. In your case I believe that is +4. That narrows it down to A, C, and E. The brute force method is to solve the standard equation when y = 0. You know that when y = 0, that x must equal \pm 6. Therefore you can check equations A, C, and E...