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?
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 =...
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...
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?