Tag Info

0

Other answers & comments cover the difference in acceleration due to friction, which will be the largest effect, but don't forget that if you are in an atmosphere there will also be buoyancy to consider. The buoyancy provides an additional upward force on the balls that is equal to the weight of the displaced air. As it is the same force on each ball, ...

5

Ball 1 will drop faster in air, but both balls will drop at the same speed in vacuum. In vacuum, there is only the gravitational force on each ball. That force is proportional to mass. The accelleration of a object due to a force is inversely proportional to its mass, so the mass cancels out. Each ball will accellerate the same, which is the ...

0

In order to show this, figure out how the transformations $C, P, T$ act on each element in the equation individually. Pay special attention to $C$. That's a tricky one! For a start on C you can check out this physics SE post.

10

I am sorry to say, but your colleague is right. Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity, $$m g = k v^2$$ And thus $$v=\sqrt{\frac{mg}{k}}$$ So, the terminal velocity of a ball 10 times as ...

0

Torque is a rotational analouge of force.It plays the same role in rotational dynamics as force playes in linear motion.Torque is defined as the moment of force or turning effect of force about the given axis or point.It is measured as the cross product of position and force vector while as angular momentum is the rotational analouge of linear momentum.It is ...

1

If you look at the two particles in the centre of mass frame you'll see something like: Your task is to find $b$. In the COM frame both particles move in an effective central potential, and the two trajectories are symmetric. I'm going to ignore the green one and just show how to calculate $b$ for the red trajectory. The closest approach is then $2b$. In ...

2

In the special case where $\ell^2$ or $s^2$ has eigenvalue zero, then $j^2$ is fixed. Otherwise you must know the projections $m_\ell,m_s$ to find $j$.

2

There seems to be no way to proceed unless more information is given. In fact, from what you have above (correcting the typo pointed out by Bernhard and ticster): $$\vec{j}=\vec{l}+\vec{s} \quad \Rightarrow \quad \vec{j}^2=\big(\vec{l}+\vec{s}\big)^2 = \vec{l}^2+\vec{s}^2 + 2\, \vec{l}\cdot\vec{s}\,,$$ meaning that knowledge of the eigenvalues of ...

3

Actually, in your rotating torus your are mimicking gravity, which is point outward. I.e. the outside of the torus acts as a floor. The centrifugal acceleration would be $g\approx\omega^2 R$. This $g$ plays the same role as the gravitational acceleration on liquid or gas pressures under normal gravity conditions, so you can say $\Delta p = \rho g h$, or in ...

0

I'm going to just use $32^∘$ below; it doesn't make a difference. Your equation isn't correct. You should have $F_x−f_k = 0$ or $Fcos32∘−f_k = 0$. The x-component of F is $F_x = F.cos32^∘$. Writing $F_x.cos32^∘$ doesn't make logical sense. Why? Shouldn't I be able to use $∑F=0$ in this problem to find the answer? Yes, you can.

0

Almost there. Since this looks like a homework question I'll just give a hint. The force balance equation is due to the buoyancy force which is due to the difference in density inside the ballon vs. outside the ballon. Once you work that out, the weight that you missing will come back into the equation.

0

So one detail I omitted from the question was that: $$\psi_{sc}(k)=\frac{g+g I}{2\pi(k^2-p^2)}$$ Where: $$I=\int^{\infty}_{-\infty}\psi(q)dq \space\space\space\space (1)$$ (I had used in arbitrary prescription in the original description of the problem, this is what I obtain before solving for $I$)$$\\$$ Using equation (1) we can solve for I, obtaining: ...

0

The air inside the balloon is less dense than the air outside; this difference is what causes the lift for the balloon. When you heat the air in the balloon, it expands until the balloon is full. At this point the balloon is still on the ground since there is not enough lift. You need to heat the air more, which expands the air more and causes some of the ...

0

As you state in the comments, $$\frac{dF}{dt}=\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t}$$ So popping this into the Lagrangian, $$L'=L+\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t}$$ The Hamiltonian $H=p\dot q-L$ implies $$H'=p'\dot{q}-L'=p\dot q+something\tag{1}$$ where $something$ is for you to work out. ...

1

OK, I will try to say something a bit more useful than my other response. I am still on the skeptical side of getting a closed-form, simple solution to this problem, specially using separation of variables. I think that the problem is that the potential outside of a square with that border conditions cannot be attacked using separation of variables. My ...

0

Using the Gauss's law (see this Link), the solution is as follow, $$\Phi (r) = \left\{ {\begin{array}{*{20}{c}} {\frac{q}{{4\pi \varepsilon {r_1}}}\,\,\,\,for\,\,\,r \le {r_1}}\\ {\frac{q}{{4\pi \varepsilon r}}\,\,\,\,for\,\,\,{r_1} \le r \le {r_2}}\\ {0\,\,\,\,\,\,\,for\,\,\,r \ge {r_2}} \end{array}} \right.$$

2

The proof is most easiest if we use the vector notation. We have $$\vec \tau = \int {d\vec \tau } = \int {(\vec r \times dm\vec g)} = \left( {\int {\vec r dm} } \right) \times \vec g$$ where I have used the assumption that near the earth $\vec g$ is constant. Now according to the definition of center of mass we have, $${{\vec r}_{cm}} = \frac{1}{M}\int ... 0 Although Logan wrote a correct solution, I find it is almost always helpful to draw yourself a diagram: Vector \vec v is the velocity, and \vec B is the magnetic field. You want to get \vec p, which requires you to add \vec m to \vec v. Thus the problem is reduced to finding \vec m, which is the projection of \vec v onto \vec B. This can ... 2 If we ignore the inner section, we have a box with 3 sides held at V = 0 and the top edge at V = V1. I'm pretty sure this is easily solvable by separation of variables using an oscillatory solution in x with a decaying solution along y. Using superposition we can then treat the inner box as a separate problem of similar geometry/boundary conditions. The ... 0 You have quite a discontinuity in the potential in two of the corners of each of the squares. I am not speaking only of the shape, but the value of the potential is discontinuous. We usually assume that the fields can have discontinuities, but the potential is always continuous. I suspect there is no solution. 1 A vector u^a is parallel-transported along the integral curve of tangent vector V^a \equiv \frac{dX^a}{d s} if we have V^a\nabla_a u^b =0 (a vector parallel-transported with respect to itself defines a geodesic). The angle between two vectors v^a and u^a is given by \cos\theta = \frac{v^a u_a}{\sqrt{v^av_a u^bu_b}}, hence we take:$$ V^a \nabla_a ...

-2

We take positive charge as a test charge because positive charge is higher potential and negative charge is lower potential. Therefore, influence of positive charge on other charges is greater than negative charges. We can also take negative charge but the effect will be lower.

1

This is just a vector projection problem. The fact that one of the vectors happens to represent a magnetic field is irrelevant. The component of velocity perpendicular to the magnetic field is the total velocity minus the component parallel to the magnetic field. I.e., it's the velocity minus it's projection in the direction of the magnetic field (also ...

2

The time dilation due to motion in a circle, relative to an observer at the centre, is just the usual Lorentz time dilation due to the velocity of the motion. If you're interested, in my answer to Is gravitational time dilation fundamentally different than other forms of time dilation? I showed how this is derived from the metric. Anyhow, as you say, the ...

1

In the Newtonian framework, you just need to solve the integral $$\int{ \frac{ G \rho_{(r)}}{|\vec{r} - \vec{r}_o|^2} \frac{\vec{r} - \vec{r}_o}{|\vec{r} - \vec{r}_o|}}dV$$ for the volume in question, where $\vec{r}$ is the distance from an arbitrary reference point to the element of matter where density is $\rho_{(r)}$, and $\vec{r}_{o}$ the distance from ...

-1

Your question 1 is non-sensical. If the earth were sqeezed (I assume you really mean "compressed") to be much smaller, then where does your flat surface come from? One can imagine a small sphere, but there would be nothing "flat". As for question 2, there are two competing factors at work. First, the effect of gravity could be lowered because some of the ...

0

Moment of force is torque. The torque that an inertia provides will be proportional to it's angular acceleration. So you don't actually have enough information in your question. The angular acceleration needs to be known to tell you what torque that inertia will provide. Simply put, how fast ($\omega$) must the dc starter turn and how soon must it get to ...

0

From the fat that the units are different you have probably concluded there isn't a single "correct" answer - because it depends on how you convert one to the other. When a flywheel (with a certain moment of inertia $I$) is rotating with an angular velocity $\omega$, it has an angular momentum $L=I\omega$. The torque this can produce depends on the rate at ...

0

This question lacks one important value in order for the answer you are looking for and that is the RPM required to get the generator started. This is because you can get the world's biggest flywheel attached, but if it is moving at a snail's pace, your generator won't start anyway. Also the "starting inertia" of $0.522~kg/m²$ should really read as ...

0

I think the key to the question is considering that $U(x)$ is periodic function therefore one has to consider an appropriate domain for integral otherwise you cannot keep the sign of $E-U(x)$ positive, in general. The fact that the particle is in a bound state implies that: $$E<0,$$ then we should have $$E-U(x)\geq 0 \Rightarrow U(x)<0$$ otherwise ...

0

You're so close to the answer I'm not sure how to nudge you along without basically just giving you the answer. I'll try anyways. There are some troubling conceptual mistakes you've made in an otherwise straightforward derivation. For starters : The reason I don't set $E_0 x_0$ to zero is because there is no $1/x$ term; otherwise I'd be able to make the ...

1

Differentiating the whole integral with respect to $f$ shows that it is strictly monotone decreasing in $f$ for $x_1<x_2$. Therefore $f=0$ is the only solution. $$\frac{d}{df}\int_{x_1}^{x_2} \frac{\sin x dx}{\sqrt{E - U(x)}}=-\frac{1}{2}\int_{x_1}^{x_2} \frac{\sin^2 x dx}{\sqrt{E - U(x)}^3}<0$$

0

It is easy if you draw the Free body diagram of the box. The FBD should show only the EXTERNAL forces. There are 2 external forces. First, the force of the earth's gravity ON the box which equals Mg and acts in a downward direction (here g = +ve). Second the force of the table ON the box which is N acting upwards. Now you decide on your sign ...

0

In "The Scalar-Tensor Theory of Gravitation", of Yasunori Fujii and Kei-ichi Maeda you can find explicitly the solution, in Appendix C (pag. 195). Personally, I really didn't like this book and even this demonstration it's very difficult to follow. So I did it in another way. Use the usual theory for the GR part, and isolate this term: $\int d^4 x ... 0 So the result is that for any given vector v in V and for any given set$S = (v_1,v_2...)$of basis states, there is a unique way of writing v in terms of the set S. The vector v is commonly used to represent a quantum state. The square amplitudes of the$v_i$in v represent the probability of seeing$v_i$when you do a measurement in the state represented ... 2 No. the whole weight will not act on the base of the container 2. If the whole weight had acted on the base of container 2, then the pressure on the base of container 2 would be equal to that of container 1 i.e., mg/A, where mg is the whole weight of the fluid and A is area of the base. But as you know the pressure at a depth depends on the height of the ... 2 The basis states are usually some orthogonal quantum states (for instance they can be the eigenstates of a Hermitian operator representing some observable). A general quantum state is then a superposition of these basis states. And because we are representing the quantum states by vectors in Hilbert space, the language of linear algebra is a natural language ... 3 It is an axiom in most formulations of QM that the state describing te system is a vector on Hilbet's space. So do not swet about it unless you wanna change the axioms (which will not be a bad idea) 5 No, the entire weight will not directly rest on the base of the slanted container (although it does indirectly). There are a number of ways to approach this, but the easiest way is to observe that the total force acting on the bottom of the container is equal to the sum of the hydrostatic pressure force (the pressure at the bottom of the container multiplied ... 1 As Rahulgarg mentioned, the pressure does not depend on the shape but on the depth. However, the direction of the force caused by pressure can be approximated as being normal to the surface, hence the total force on the sides will depend on the shape. For a fluid at rest like the one I think you are assuming, the pressure at any depth will be$p=p_0+\rho g ...

0

This may be cheating, but I think the problem is easier if you use conservation of energy. If you set the gravitational potential energy reference to the height of m1, then initially you have $$U_i = k\frac{q^2}{d}.$$ The final energy will have a gravitational potential energy and an electrical potential energy: $$U_f = mgh + k\frac{q^2}{r}$$ A bit of ...

0

I worked the problem out a ways and it involves quite a bit of tedious algebra. The technique I used was to include the electric force in the Fx and Fy equations by looking at the angle that $\hat{r}$ (the vector between the two charges) makes between the charges. For example, the equation i came up for Fx is $T_x - \frac{Kq^2}{r^2} \cos{\theta} =0$ where ...

0

eVs= hv + work function of the metal. So putting e= 1.6*10^-19 and h stands for planck's constant=6.62606957 × 10-34 m2 kg / s v= frequency and simply hv as the energy of the photon.

0

For your second problem, the propagator can be written with its indices as $$(S_F)_{\alpha\beta}(x-y) = \langle T\{\psi_{\alpha}(x)\bar\psi_{\beta}(y)\} \rangle$$ Then we have $$\langle T\{\bar\psi(x)\Gamma\psi(y)\} \rangle = \Gamma_{\alpha\beta} \langle T\{\bar\psi_{\alpha}(x)\psi_{\beta}(y)\} \rangle = -\Gamma_{\alpha\beta} ... 1 If you replace the fluid scenarios with solids with matching have the same shapes, then indeed the pressure at bottom would be the weight of each solid divided by base area, i.e., they would be different! Then what's different about the fluid case? I presume in scenario 1 you have a prism, in scenario 2 you have a bottom-heavy frustum, and in scenario 3 ... 1 The efficiency of large hydroelectric generators can be very high - up to 95% in ideal cases - however the efficiency of small installations is a lot lower and in particular it's hard to get efficient electricity generation if the flow rate is low, which is likely to be the case for your friend. But lets see what the potential is. You don't say what the ... 2 Consider this diagram showing the three columns you describe all connected to the same body of water: Your question asks whether the three pressures P_1, P_2 and P_3 will be the same. The answer is obviously yes, because the columns are all connected to the same body of water. For example if P_1 > P_2 then water would flow from the base of ... 0 The current I has a value in one point of the circuit, in contrast to the voltage U which is always measured between 2 points. The definition of I is the amount of charge \Delta q that passes through a particular point in the circuit in the time \Delta t (it's a quantity mathematically similar to the simple velocity in kinematics). So when you ... 0 There is a flaw in Step 2. The formula should read,$$(\frac{\Delta C}{C}) = (\frac{\Delta m}{m}) + 2(\frac{\Delta d}{d}) + (\frac{\Delta h}{h})  I still don't see how you arrived at your formula. It is completely wrong. Though it looks close to the formula for variance. Maybe that's where you got it confused. Conceptually, for $f(x,y,z,...)$ ...

2

The first bullet is correct, the outer shell does not contribute. This easily follows from Gauss' law. For this you use the fact that the electric field must be radial and any cylinder inside the cylindrical shell does not enclose the charge density $-\lambda$. You might think that close to the negatively charged shell there is an additional electric field ...

Top 50 recent answers are included