Recent content by Avalanche

C = (1/4+1/11)^-1 = 2.9333 Time constant = RC = 5*2.93333 = 14.6667 microseconds 0.5 = e^(-t/14.6667) t = 10.2 microseconds Thanks for all the help! btw ... does it also take 10.2 microseconds for the potential across the 6 microfarad resistor to drop to half of its initial value?

The time constant = RC. The only resistor is the 5 ohm resistor. Is that R? And for C, do I use the equivalent capacitance or just the 4 microfarad?

No ... it's the right loop. When the capacitor discharges, it creates a current that flows around the right loop. I3 becomes zero. Is that correct?

When the circuit is opened, the left loop should still has current because there is a battery in that loop.

Homework Statement Please look at the attachment for the circuit. The question is this: The switch has been closed for a long time. It is then opened. How long does it take for the potential across the 4μF capacitor to fall to half of its initial value? Homework Equations V =...

I guess my question was why the equivalent resistance wasn't this

I think I kind of get it. Why do you have to rearrange the circuit? Is it because you want the resistors to be in parallel with the capacitor?

Homework Statement Please look at the attachment. The part I am having problems with is part C Homework Equations time constant = RC Q = Q0 e^(-t/RC) The Attempt at a Solution I don't understand why the circuit is equivalent to the 4 ohm resister with the 18 ohm resistor in series so that...

Looking at the answer, I assume springs in parallel have greater spring constants, then springs by itself and springs in series have the smallest spring constant. Is there a reason for this? Like an equation/relationship?

Greater the mass, greater the spring constant. Longer the spring stretches, smaller the spring constant. Since C stretches by the largest amount and has a mass of only m, the spring constant is the smallest. A and B stretches by the same amount and have the same mass, the spring constants are...

Homework Statement Homework Equations Hooke's law. F = kΔx ω = sqrt(k/m) The Attempt at a Solution For part A F = mg = kΔx k = mg/Δx g is a constant so the spring constant is proportional to the mass and inversely proportional to the change in distance that the spring...

So the mass of the board= density of water*volume of water displaced Because the object is floating, the volume of the water displaced is equal to the volume of the submerged part. so volume of water displaced = vo = 0.12 m^3 = 1000*0.12 =120 kg density of boat = m/v =120/0.36 = 333.3 kg/m^3...

I need the mass of the object. How do I find that? Is rhoo equal to the density of water?

Homework Statement Homework Equations density = mass/volume FB = rhoo vog for a floating object F = mg The Attempt at a Solution Buoyant force = force of gravity due to the board rhoo vo = m vo = area*height submerged vo = 2.4*.05 vo = 0.12 m^3 volume of the board =...

So for this question, the reference height of 0 is already set at the manometer? Is it possible to use the reference height at the beer level in the container?