Recent content by GrandMaster87

Homework Statement Hi Guys I have a problem when working with conservation of energy and work energy theorem. I have a problem when friction is present in the system. I don't know when to add the friction to get the total work done or when to minus the friction to find work done. Will...

hi...i just get a bit confused on when to use the brackets and commas in a name or not...

so when the emf is negative and the electrode is connected to the positive side it is the anode thanks a lot guys for your help.

thanks a lot man! I am still a bit confused by i will make my way to understand it :) will it be safe to say if an electrode is connected to the positive terminal of the voltmeter it will be the cathode?

Hi...has anyone seen this thread as yet?

Homework Statement Two learners, Wade and Tang are working together on a project. They were asked by their teacher to assemble a galvanic cell that would produce a large enough voltage to light up a 1.5V bulb. They are allowed to use two electrodes, one of which must be a zinc electrode...

Homework Statement Solve for x using general solution sin2x = 0.5 Homework Equations sin2x= 2sinxcosx The Attempt at a Solution arcsin0.5 = 2x 2x = 30 degrees + k360 x = 15 degrees + k180. Or 2x = (180-30) + k360 2x = 150 + k360 x = 75 + k180

i did an i get 0.37s...and if the the second stone was released 1 s later then how can that be?

Homework Statement A mountain climber stands at the top of a 50m cliff that overhangs a calm pool. She throws two stones vertically download 1s apart and observe that the two stones reach the water simultaneously after a while. The first stone was thrown at an initial speed of 2m^{-1}.s...

can we expand sin(3x) and cos(3x) using double angle formula?

thats as far i got...im really new with trig ..caught a wake up call at school so i started working with it...

Homework Statement Prove that \frac{sin3x}{sinx}-\frac{cos3x}{cosx} = 2Homework Equations The Attempt at a Solution LHS:\frac{sin(2x+x)}{sin}-\frac{cos(2x+x)}{cosx} =\frac{sin2x.cosx + cos2x.sinx}{sin}-\frac{cos2xcosx - sin2x.sinx}{cosx}

1. thanks a lot :) forum is quite cool..im gng to study now...and will come back if i need more help thank you a lot Doc Al

Wont cosx be Cos0 ?

Ok...basically this is what i have done so far. Fgravity=mg(sinx) =50 x 9.8 x sin25 = 207.08N Ffriction=60N Therefore force parallel =207.08-60 =147.08 But they also tell us to work out work done we must use F(cosx)\Deltax so i want to know...when and why are we using sinx...