Let's start with the 1uF capacitor. This is in parallel with the 3 ohm resistor, so the voltage across the capacitor will be the same as the voltage across the resistor. Yes?
So we have a 3 ohm and 6 ohm resistor in series with 9V across the two of them. That means 3V is dropped across the 3 ohm resistor and 6V across the 6 ohm resistor. Yes?
@Scáthach The same argument applies to the 3 uF capacitor. It is in parallel with the 7 ohm resistor so the voltage will be the same as the voltage across the 7 ohm resistor.
We generally write the equation for radioactive decay as:
$$ \frac{dN}{dt} = -k N $$
where $k$ is the decay constant. I guess what the question means is that $k = A + B$, where $A$ is the constant for the electron capture and $B$ is the constant for beta minus decay.
@JohnRennie Sure, I'm looking at some Feynman problems and I've encountered this:
Two gliders are free to move on a horizontal air track. One is stationary and the other collides with it perfectly elastically. They rebound with equal and opposite velocities. What is the ratio of their masses?
Apparently it's a specific number from the answer key... I don't see how.
Alright, well in a perfectly elastic collision kinetic energy is conserved so $\frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1'v_1'^2 + \frac{1}{2}m_2'v_2'^2$ and similarly for momentum, $m_2v_2 = m_1'v_1' - m_2'v_2'.$
don't know why I wrote primes on the masses, but we'll go with it
Call the initially moving mass $m_1$ and its velocity $v_1$. After the collision $m_1$ has the velocity $-v_2$ and $m_2$ has the velocity $v_2$. Then from momentum conservation we get:
Well let's have a look. First let's take a side view in the plane that is lined up with the meridian i.e. the plane is lined up with the North and South poles:
My diagram shows the dip in the plane of the meridian, but the question says the compass is being held in a plane at 30° to this. Let me draw a bird's eye view of this i.e. looking from above down onto the surface of the Earth:
@Scáthach Does this make sense. I can attempt to draw it in 3D if that would help.
The compass is being held in a vertical plane at 30° to the meridian, and the blue line shows this plane (remember we are viewing this from above so we see the planes edge on)
@Abcd The pulley B isn't fixed, but in the rest frame of B, i.e. the frame in which B is fixed at the origin, we still require that the accelerations of M and m be equal and opposite. So let's call these accelerations $+a$ for $m$ and $-a$ for $M$. OK so far?
But the pulley B is accelerating i.e. the rest frame of B is an accelerating frame. If the pulley B is accelerating at $+A$ then to get the accelerations of $m$ and $M$ we need to add $+A$ to those accelerations.
So in the lab frame the acceleration of $m$ is $+a + A$ and the acceleration of $B$ is $-a + A$.
So in the lab frame the accelerations of m and M are not equal and opposite.
No because the force on the pulley B depends on the tension in the string joining m and M, and that depends on how fast m and M are accelerating.
But this is still just a matter of writing down all the forces, tensions etc and grinding through the equations. This type of system is called an Atwood machine if you want to Google for how to approach these problems.
@Scáthach If you know the linear charge density, in coulombs per metre, of the ring then you can multiply this by the linear velocity, in metres per second, of the ring and that gives you current, in coulombs per second.
@JohnRennie A balloon of mass M with a light rope having mokey on the rope is in equilibrium. if the mokey starts moving up with accln a wrt the rope, then the accln of the centre of mass of the system is?
@Scáthach If you know the charge per unit length, $\rho$, then you can multiply it by the length $\ell$ to get the total charge corresponding to the length $\ell$ i.e. $Q = \rho\ell$.
But the question is deliberately obscure. For example it doesn't give you the radius of the ring. But if you write the radius as $r$ and do the calcualtion you should find the $r$s cancel out in the end to give you the answer as a number.
@Scáthach We can go through the calculation if you want ...
The net external force on the balloon/monkey system is zero because the upthrust due to the buoyant force is equal to the weight of the balloon and monkey. And without an external force the position of the COM can't change.
The left side experiences a force up out of the page and the right side experiences a force down into the page, so yes that makes the direction of the torque upwards in the plane of the page.
@Scáthach I have to go now I'm afraid. I'll be back later today or failing that tomorrow morning.
If the capacitor charges or discharges instantaneously that means some charge $Q$ flows on or off the capacitor in zero time. So the current is $I = Q/0 = \infty$.
@pi-π OK, so you know that if you reverse bias a (non-zener) diode no current flows. Then if you keep increasing the reverse bias voltage at some point you get an avalanche breakdown and suddenly a huge current flows and the diode usually burns out. Yes?
@pi-π zener diodes actually use two different mechanisms. In low voltage zener diodes the electrons quantum tunnel across the depletion zone at the zener voltage. Strictly speaking the term zener only applies to this mechanism.
However this doesn't work well for higher voltages.
So higher voltage zener dioides are designed to undergo an avalanche breakdown at the diode voltage, but they are designed so the breakdown is controllable, unlike in a conventional diode.
I usually don't distinguish between the two breakdown modes because the end result is the same.
@Abcd I'm not sure what the significance of the plank is. As far as I can see it's just a smooth hemisphere on a smooth surface where smooth presumably means frictionless. In that case the centre of mass of the hemisphere cannot move sideways - only up and down.
Suppose you stuck a pin through the centre of mass of the hemisphere (the point C in the diagram) so the hemisphere rotated about the point C. In that case the centre of the sphere would rotate in a circle about the point C.
But C moves up and down as the hemisphere slides on the smooth surface, so the path traced out by the centre of the sphere cannot be a circle.
