Recent content by SuperHero

oh sorry i re-did this this way now: Ib= V/R Rtot = R + 2R Rtot = 3R Ib = V/3R for Ia = V/R Rseries = 5R Rparallel = 6/2R Rtot = 10R^2 + 6 / 2R Ia = V/20R^3 + 12R right?

Homework Statement Part of an elecrial circuit for the olympic score board consisted of a single battery V and several resistors having on of two values. R or 2R. They are hooked up in the confriguration as shown. Calculate the ratio of the two currents shown Ib/Ia...

thx for your help, i will look over.

M = 2m(2x-3x2) = 2my M = 2(30kg)(1/3) M = 20kg YES THE MASS IS 20KG! :D

but isn't y the mass ?

y = -3x^2 + 2x y = -3(1/3)^2 + 2(1/3) y = -3/9 + 2/3 y = 0.3333

x = 1/3 since the greatest y is at the maximum

down :)

since i found the x = 2/3 can i do cos θ = 2/3

it is a parabola..

so i should use the quadratic formula y = -3x^2 + 2x x = -(2) +/- √ ((2^2)-4(-3)(0)) / 2(-3) x = -(2) + 2 / -6 x = 0 or x = -(2) - 2 / -6 x = -4/-6 x = 2/3

M = 4mcosθ - 6(cosθ^2)m M = 4mx - 6(x^2)m

T = Fg - 2mg(2-3cos(θ))cos(θ) 0 = Fg - 2mg(2-3cos(θ))cos(θ) Fg = 2mg(2-3cos(θ))cos(θ) Mg = (4mg - 6cosθmg) cosθ Mg = 4mgcosθ - 6(cosθ^2)mg M = (4mgcosθ - 6(cosθ^2)mg)/g M = 4mcosθ - 6(cosθ^2)m

but see i do not know the angle which is really hard

T = Fg - 2mg(2-3cos(θ))cos(θ) (A cos(θ) + B)2 + C. T = -mg(-3cosθ +2)^2 + Fg like this?