How do I solve this problem? The period of the free oscillations of the system shown here if mass M1 is pulled down a little and force constant of the spring is k and masses of the fixed pulleys are negligible, is
The answer is 2π √(M2+4M1/k)
Here M2 is mass of movable pulley and M1 is mass of the block
I deduced that when M1 moves down by x, M2 moves up by x/2 and therefore spring stretches by x/2,but I'm not sure what to do next.
The vapour pressure is the pressure at which the chemical potential of the component in the vapour and the solution is the same, so you'd have to consider how the chemical potential varied in both phases.
If you were doing a chemistry degree then this would be a very interesting thing to pursue because it would teach you lots about thermodynamics. However I think it would be a somewhat pointless diversion right now.
@JohnRennie A plane spiral with a great number N of turns wound tightly to one another is located in a uniform magnetic field perpendicular to the spiral's plane. The outside radius of the spiral's turns is a. The magnetic induction varies with time as $B = B_O \sin (\omega t)$, where $B_o$ and $\omega$ are constants. Find the amplitude of induced emf.
Suppose we had a long straight wire of length $\ell$ with a field $B$ normal to it, then the EMF developed between the ends of the wire would be $d/dt ( B \ell )$. Yes?
I can see what the solution is doing. If you take a loop of radius $r$ then the flux linked is $\pi r^2 B$ and $d/dt$ of this gives you the EMF. You treat the spiral as being made up of those loops and integrate to sum up the EMFs.
(approximating the sum as an integral because $N$ is large)
I'm just not sure I understand why that is a valid treatment.
@Abcd I'd guess it means the radius of the ring is only just big enough for the ring to fit round the solenoid. The question is probably going to assume the flux through the ring is the same as the flux through the core of the solenoid.
@JohnRennie A long straight solenoid of cross sectional diameter d= 5cm and with n = 20 turns per cm has a round turn of copper wire tightly put on its winding. The cross sectional area of copper turn is $S= 1mm^2$
@JohnRennie Just look at the cross sectional area!^^ as compared to the area of solenoid!!
The force between them acts at the contact point and normal to their surface, so it acts at an angle as shown by the red arrows. That means the force pushes the disks sideways.
The calculation isn't as hard as you think. Momentum is always conserved so the total vertical components of the momenta must add up to zero as well as the total horizontal components.
And energy is conserved, or if the coefficient of restitution isn't unity you can still calculate what the final energy is.
The hard bit is that for an oblique collision the momentum change, the impulse, is a vector that lies along the line connecting the two centres of the disks.
Anonymous
@Abcd Why not? We can divide a disc into infinitesimal loops and calculate the emf due to each and add them up....which is basically what they're doing while integrating. For sure, they're assuming that the wires aren't insulated.
Yea so the momentum conservation fails there right
Anonymous
@AvnishKabaj Not a great comparison because wires and discs can be made of the same material (say copper), whereas pizza and chocolate are not. :P
Anonymous
Also, current/voltage is due to free electrons. So you don't need to worry about whether the consecutive loops are physically "joined" or not. Just contact suffices.
I deliberately drew the diagram so at the moment of the collision the line joining the centres of the disks was at 45º.
That means the impulse has equal horizontal and vertical components.
i.e. the change in the horizontal component of the momentum is equal to the change in the vertical component of the momentum.
In your problem with the big disk and the two little disks you'd have to work out the angle of the impulse vector. It's not that hard - it's just geometry - but adds an extra complication.
I need to work I'm afraid so I'll have to call it a day now.
I got it. Basically the impulse is 45 with the line of collision and thus will have equal components along and perpendicular to the line of collision
And impulse is change of momentum so the changes will be equal in x and y components
Anonymous
A plane spiral with a great number N of turns wound tightly to one another is located in a uniform magnetic field perpendicular to the spiral's plane. The outside radius of the spiral's turns is a. The magnetic induction varies with time as $B = B_O \sin (\omega t)$, where $B_o$ and $\omega$ are constants. Find the amplitude of induced emf.
Anonymous
I thought we were talking about this question @AvnishKabaj
@AvnishKabaj It's a plane spiral made of only one wire. And the consecutive loops closely touch each other.
Anonymous
Be careful though as they're using two different notions of loop in the question. They're actually integrating over infinitesimal rings of constant radii.