Recent content by dan greig

sorry that was a bit of a marathon, thank you for your help

sorry it was supposed to be 2x^3-5x^2/2-6x+c so that gives me, c = -3-16+10+12 = 3

would my final equation be, y = 2x^2 - 5x^2/2 - 6x + 3 ?? i also need to sketch the graph, to find the points where the line intercects the axis do i use, y = 0 for x axis x = 0 for y axis

do you mean 6x^2 - 5x -6 the result of this using x = 2 gives 8 does this mean, -3 - c = 8 adding 3, -c = 11, c = -11 ? sorry to keep on but i need to get this correct!

if i expand the brackets from (2x-3)(3x+2) i get, 6x^2 + 4x - 9x - 6 and then integrate, y = 2x^3 + 2x^2 - 9x^2/2 - 6x +c do i then just plug in the x and y values to find c by rearranging, y - c = 2x^3 + 2x^2 - 9x^2/2 - 6x ?

if i have calculated the charge and energy stored in the capacitor can i then use, energy dissipated = 1/2 QE to find the energy lost as heat, therefore the energy needed to charge the battery to 24v is double the energy stored? or would the energy needed be, the energy stored plus the...

how do i calculate the energy dissipated?

so is there an equation to find this?

log(10^x) = x ? analogous?? The same as?? but log e = 1 does that mean log e^x = 1^x ? therefore log e^x = x ?

I have a question that has confused me a little, A 10 micro-farad capacitor is fully charged by a 24v battery. The question asks how much charge is stored and how much energy would be stored. I think i have found the answer to these using, Q=CV and E = 1/2 CV^2 respectively but...

ln(e) + x ? not really sure you've lost me a bit sorry

would it go to, V = Vo ln + (-t/RC) then to, V = ln Vo - t x 1/RC

how would i rearrange this equation to find the capacitance, c? V=Vo exp(-t/RC)

so i get, (x^2 - 3x)(3x^2/2 +2x) ?

so if i differentiate i get, 6x^2-9x-6 and then integrate, 2x^3-9x^2/2-6x ?