Recent content by SagarPatil

I have found the solution to the answer. Thanks for helping everyone

So I found the expression for dW/dR After the deravation, if I set dW/dR to 0, and set r and E to 1, I get a L shape graph. The x and y basically start from origin and y is + and x is positive

So, I was wondering, When you find the expression of dW/dR, do you need to find the second derivation of it to graph it or just find the fist derivation of dW/dR and set it to 0.

I find product of two expression harder to do. So I derived for dW/DR and this is what I got If I set R = 0 I would just get 0.

ok, so I am using the relative velocity to find VD. The red arrow for vd is Vd/B Vd is at point D going left an angle of 45 degree (assuming)

Looks right to me Ah, I see I didst use the correct radius. Thanks! Yea, I will put the direction in the end

Homework Statement Velocity of point C = 0.2 m/s going down Find VB, VD, Angular velocity of BCD and Angular Velocity of AB Homework Equations V=wr Sine and cosine law The Attempt at a Solution Drawing of the question Please note that BCD is one link (L shape)

Okay, I have done some work. I found the derivative of W'(R) and set it to 0. Here is what I got

Homework Statement So a generation with an internal resistance r delivers a voltage E. The generator is connected to a electrical circuit with a resistance R. The work W done each second in sending a current through a circuit with a resistance R is given by W =E^2*R/(R+r)^2. For constant r...

Homework Statement Dont really need the answer, but I want to know if point C has Cx and Cy or just Cy. I am confused there.

I got the answer. I didn't understand the concept first. After reading I got it :D

If I take moment at point A does Bx cancel ? BC I calculated is correct?

1. The problem statement, Have to find the forces in AC & BC using the methods of section. all variables and given/known data Distance AB = 79cm BC = 63 AC = 98 d = 3.833 AZ = 62.881 ZC = 75.16 Point B is a roller and point A is a Pin Whats given is 39.94N Forces @ C going down Homework...

Hello, I got it. You so here is what I did Area of the Big sector = 5.270^2 * 0.97(rad) / 2 = 13.5 Area of Small Sector = 0.625^2 * 0.97(rad) / 2 = 0.189 Area of the big circle = 1.23 Area of small circle = 0.2 So, 13.5+1.23-0.2-0.189 = 14.341 14.341 * 0.250 = 3.58 in^3

is the answer correct ?