Recent content by xSnoopy

Okay I seem to got the first part Torque on Wheel = Torque on coils in motor TorqueWheel = ForceFriction*RadiusWheel TorqueCoil = B*I*(AreaCoil) Where I can solve for I or current. :) But now I'm stuck on calculating the back emf... I guess I know... V(battery) - IR = emf or -L(dI/dt) =...

Homework Statement A small electric car overcomes a 269N friction force when traveling 28.6km/hr. The electric motor is powered by ten 12.0V batteries connected in series and is coupled directly to the wheels whose diameters are 50.6cm. The 295 armature coils are rectangular, 10.0cm by 14.8cm...

Homework Statement a)Calculate the work performed by an external agent to move a charge of -0.39x10^-12 C from `i' to `b' http://capa-new.colorado.edu/msuphysicslib/Graphics/Gtype54/prob04a_threeqcontour.gif Hint: The equipotential lines shown are separated by 1 kV. Work to move a charge is...

You get μ(2.94) + (2.94) = 0.6a ?

So you would get T-μ(0.3*9.81) = 0.3a (0.3*9.81) - T = 0.3a and substitute ?.. T-μ(0.3*9.81) = (0.3*9.81) - T that doesn't cancel out the T's ?

Oh like net force? The vertical would then be.. T - 0.3a = 0.3a ?

so Force normal = 2.94N and force moving horizontal would be (0.3)(a) then μ(2.94) + (0.3)(a) = (0.3)(a) ? ... crap I'm not getting something here :/

Well its not at equilibrium so i can't just sum the forces to get zero ...?

I need help with my homework. http://img36.imageshack.us/img36/8717/friction.png Uploaded with ImageShack.us I don't how to figure this out because you only get velocity :/