Gas A: Cp = 29 SI unit and CV= 22 SI unit Gas B: Cp = 30 SI units and CV = 21 SI units A) Both have one mode of vibration B) A is rigid and B has mode of vibration C) B has no mode of vibration and A has one D) A has one mode of vibration and B has two Both gases are diatomic given
@PolarBear Well... I just wrote Cv for each gas as (5+x)R/2 and Cp as (7+x)R/2,equated them to their values and divided the equations.... For A i got a consistent solution of x just greater than 1..while for B i got a negative value of x... So C
The thing is that both A and B have specific heats that are a close match for a diatomic at a temperature where no vibrational modes are active. Each vibrational mode should add R to the specific heat.
But there is no option saying no vibrational modes for both.
This was the problem so ; I thought cross section is a function of x . Ie A(x) = a+ kx . A= b when x= l so $k = b-a/l$ now i wanted to find the resistance via $\frac{\rho l}{\pi}\int \frac{dx}{(a+kx)^2}$ where x limit goes to 0 to l. And we finally come to result R = $\frac {\rho l^2(b-a)}{\pi ...
@JohnRennie forget this . Thats done. I can't believe that i was doing a wrong integration till yesterday being an undergrad. Anyway do you have anymore free time?
@Nobodyrecognizeable that's a system of two coupled oscillators. There will be two normal modes, one in which the masses move in phase and one in which they move in antiphase.
Suppose $x_1$ is the displacement of the upper mass and $x_2$ is the displacement of the lower mass then the tension in the upper spring is $T_1 = kx_1$, and the tension in the lower spring is $T_2 = k(x_2 - x_1)$
@JohnRennie hey again. quick question: consider a toy drawbridge (I will attach the picture right after) of mass $m=4kg$ and $l=30cm.$ The question is:What is the angular velocity of the bridge when it falls to the point that it is hanging vertically?
I realize this is quite easy to solve with conservation of energy, but I was wondering why another method using the work-energy theorem with $\int_{40}^{0} \tau d\theta = \int_{40}^{0} mg\frac{l}{2}\sin(\theta) d\theta$ so that that integral $= \frac{1}{2}{\omega}^2$ doesn't seem to work
@Abcd See if the battery is not connected then charge don't have any place to go , and since it can't be destroyed it stays there , whereas if a battery is connected then new charge can be moved in /out
@Mr.Xcoder you have nationals . So probably you should take up some problems. I suggest you walter lewin's problems in youtube .they are worth giving a try.
@Nobodyrecognizeable Thanks for the suggestion, I will have a look. I do know Lewin from some online lectures I've watched a while ago, I think. Right now I am preparing for the experimental examination and I don't have much work to do while I'm on my way home :P
@pi-π still no. If the electron is moving parallel to the electric field lines it will move in a straight line, though it will be accelerating not moving at constant velocity.
@JohnRennie before immersion of the bob, the weight of the tumbler was $10M$ where $M$ is the mass of the water and the tumbler is assumed to be mass-less. The mass of the bob is $m$ and is fully submerged in water.
@JohnRennie I think the mass should remain the same because the bob is not suspended in the water. Its held by a rigid body
I mean its suspended but its not freely floating in the water
Suppose the bob was something very light, like a ping pong ball. You put the ball on the top of the water, so it's floating on the water, and then the weight shown on the scales goes up by $mg$. Yes?
Now you press down on the ping pong ball with your finger (which is a an approximately rigid object) and the weight shown on the scales will increase by an amount equal and opposite to the force you're pressing with.
@user8718165 I think the easiest way to look at this is by starting with an object that has the same density as water. When you add this object to the beaker you are just adding a mass $m$ of water. There won't be any force in the rod because obviously water has neutral buoyancy in water. So the weight on the scales goes up by $mg$.
@user8718165 Now we have already discussed what happens if we keep the volume the same but decrease the density. In this case the rod has to push down to keep the object submerged. So by Newton's third law the water/beaker/scales have to push up with an equal and opposite force. So in this case the weight shown on the scales goes up. Yes?
Hmm, yes, it would, but let's assume the rod is very thin so it has little effect i.e. the force is dominated by the object attached to the end of the rod.
@JohnRennie ok sir...got upto this...could you please tell why would the mass of the rod even count? Isn't the rod attached to an external frame and hence doesn't apply any force on the water?
@user8718165 Anything underwater exerts a force on the rod. The underwater bit will be the ball on the end of the rod and the part of the rod that's underwater. The force can be upwards if the ball/rod average density is less than water or downwards if the ball/rod average density is greater than water. OK so far?
@user8718165 if we are counting the part of the rod that is underwater, as well as the ball, then the densities of the rod and ball might be different.
For example if we put a ping pong ball on the end of a steel rod then the ball will be less dense than water but the rod will be more dense than water.
The reason I said average density is that we need to take the densities of both the rod and ball into account.
Or we can just say that if the rod is thin its effect is negligible and just ignore it!
Actually, wait, no, I've got this wrong. The density of the ball and/or rod doesn't matter. Only the volume matters.
That's because the upthrust on the submerged body is the weight of water displaced i.e. $\rho_w V g$, where $\rho_w$ is the density of the water. And the increase in the weight on the scales is equal to the upthrust.
The reason I was going on about the density of the object is because if we start from the mass $m$ then the volume is $m/\rho_o$, where $\rho_o$ is the density of the object.
If we substitute this into my expression for the upthrust we get $\frac{\rho_w}{\rho_o}mg$, so that's where the density of the object comes in.
But if we stick to working with the volume of the object then it's a lot simpler.
That's just the usual $\sigma = \eta \dot\gamma$, where $\sigma$ is the stress and $\dot\gamma$ is the strain rate.
But that's only useful when we have a constant strain rate. For a ball falling in a liquid the strain rate is high near the ball and decays rapidly to zero as we move sideways away from the ball.
For Stokes' law to be valid we just require that the walls of the container are far enough away that the strain rate is approximately zero near the walls.
In that setup you have a laminar flow in the thin layer between the falling object and the edge of the tube. If the velocity of the falling object is $v$ then the velocity of flow goes from zero at the inner surface of the tube to $v$ at the outer surface of the falling cylinder in a distance $y$, where $y$ is the size of the gap between them.
Since the flow is laminar the velocity increases linearly over the distance $y$, so the strain rate is $\dot\gamma = v/y$
And the stress is $\sigma = \eta\dot\gamma = \eta v/y$
Oh, ok. I thought both were relevant and viscous force was only due to velocity gradient of fluid between stationary surface of container and moving surface of the falling body.
@Dante no, with a falling sphere in a large container the strain rate is only non-zero in the immediate vicinity of the body and it fals quickly to zero in a non-linear way as you move away from th body.
If a projectile is launched at a speed $u$ from a height $H$ above
the horizontal axis, and air resistance is ignored, the maximum
range of the projectile is $R_{max}=\frac ug\sqrt{u^2+2gH}$, where $g$ is the
acceleration due to gravity.
The angle of projection to achieve $R_{max}$ ...
A and B will be two different capacitors, I get that and then that will be an LC circuit.
Next, how an I supposed to solve it?
I mean it would go from A to L and stay there (?) And then move on
A and B would act as capacitors and that's an LC circuit. Now, the charge would go from A to L and then to B. Now, considering that circuit we'll have \omega = 1/√2LC where C = capacitance of one sphere. So, let T be the time period. Like I don't really know what will happen, what I know is the charge will go into L get stored after time T/2. Now, it will start journey to sphere B and reach it in T/4.
If a projectile is launched at a speed $u$ from a height $H$ above
the horizontal axis, and air resistance is ignored, the maximum
range of the projectile is $R_{max}=\frac ug\sqrt{u^2+2gH}$, where $g$ is the
acceleration due to gravity.
The angle of projection to achieve $R_{max}$ ...
