# Tag Info

27

This diagram shows the Earth rotating round the Sun at it's orbital velocity $v$. That is the centre of the Earth is orbiting around the Sun at velocity $v$. NB the scale is rather fanciful - don't take it literally! I'll also assume the orbit is circular, and for convenience I'll ignore the Earth's rotation i.e. assume it's tidally locked. To calculate ...

16

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 ...

11

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 ...

10

Let's simplify. Let's eliminate the Moon. Let's get rid of the Sun temporarily. Let's replace the Earth with an equivalent mass-and-density perfect sphere of iron that is neither moving linearly nor spinning or revolving in any way. We place two 1KG iron test masses on opposite sides of the Iron Earth, suspended 1 M above the surface by identical ...

8

The mistake you're making is that you're looking at the full acceleration when you should look at the relative one. At distance $R=1\mathrm{au}$ from the sun, the gravitational acceleration is given by $$a_0 = \frac{GM_\odot}{R^2}$$ Assuming a spherical cow earth (in vacuum), at midday at the equator, we're one earth-radius $r$ closer to the sun, ie $$... 6 Yes your weight will change. The moon will have a bigger impact than the sun, so you may be heavier during the day (when there is a full moon). The effect has been measured: This figure is on page 93 of "Practical Physics" by Gordon Squires (a classical book, and one that I highly recommend). The method used is a beautiful example of careful experimental ... 5 No, the entire weight will not directly rest on the base of the slanted container (although it does indirectly). There are a number of ways to approach this, but the easiest way is to observe that the total force acting on the bottom of the container is equal to the sum of the hydrostatic pressure force (the pressure at the bottom of the container multiplied ... 4 Faraday's law fails here. Let's go back to basics. We use the Lorentz force. And what is happening is as the rod rotates, the charges in it rotate too. However, the rod is neutral so there is no net current flowing. Now field of the bar magnet is towards left in the wire, the lorentz force applies on the protons and electrons inside the wire, causes the ... 4 Assuming the tube is insulated along its whole length (which I think is the intent of the question although it's not stated), I think a flow in either direction is stable and sustaining, but there is no particular reason it will form in either direction if the initial conditions are that the air in the tube is still. It will then depend on the average air ... 4 If you draw the path, it looks like this: The person moved 10 m in 5 seconds. 4 Take the equations of motion and plug in a generic parabola. It should be easy to prove that no constant coefficients of the parabola satisfy the equations of motion. From  \vec{a} = \vec{g} - \beta |\vec{v}|^2 \frac{\vec{v}}{|\vec{v}|}  by component$$ \begin{pmatrix} \ddot{x} \\ \ddot{y} \end{pmatrix} = \begin{pmatrix} 0 \\ -g \end{pmatrix} - \beta ...

4

Following the same orbit at greater velocity would mean that the satellite has greater acceleration (its velocity is changing at a greater rate than if it follows the same route more slowly). But the only force acting on it is earth's gravity, which creates a particular acceleration at any given altitude above the earth. It can't go any faster or slower ...

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 ...

3

I think you have some problems with the sign convention. At first, ask yourself the question "why did I need the sign convention?" For the moment, forget the sign convention and let's derive the formula related to lenses from the scratch using only geometry. Let us consider: p= distance of the object from the optical centre q= distance of the object ...

3

A stationary state, usually referred to as the steady state, is the particular case when the partial time-derivative of the variable in question is zero. For your question, $$\frac{\partial v}{\partial t}=0$$ Or, in macroscopic terms, $$\frac{\Delta v}{\Delta t}=0\to v_{in}=v_{out}$$ which is what jhobbie said in the comments. The water level will ...

3

To derive the field equations more quickly, consider that all the terms in the Lagrangian are "squares": a tensor contracted with itself on all indices. The variation of a square is $\delta(F_{ab} F^{ab}) = 2 (\delta F_{ab}) F^{ab}$ in analogy with $d/dx\; f^2 = 2ff'$. Using this, we have for the variation of the Lagrangian $$\delta \mathcal L = - (\delta ... 3 You know that in GR you need a locally Minkioski spacetime. This, in each point of your manifold you can change the coordinates so that the metric is diagonal, and the square of the infinitesimal displacement is ds^2=\left(ct\right)^2-x^2-y^2-z^2. So here is where the c come from. Then, when you want to compute the coupling constant k=\frac{8\pi ... 3 It is an axiom in most formulations of QM that the state describing te system is a vector on Hilbet's space. So do not swet about it unless you wanna change the axioms (which will not be a bad idea) 2 The proof is most easiest if we use the vector notation. We have$$\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 ...

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 ...

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 ...

2

If the flow is laminar, i.e. not turbulent, then the relationship between flow rate and pressure is given by the Hagen–Poiseuille equation: $$\text{Flow rate} = \frac{\pi r^4 (P - P_0)}{8 \eta l}$$ where $r$ is the radius of the pipe or tube, $P_0$ is the fluid pressure at one end of the pipe, $P$ is the fluid pressure at the other end of the pipe, ...

2

The question defines $V$ as the rate at which the string is pulled downward through the ring and consequently the rate at the which the radius changes, $-dr/dt$ (negative as the radius is decreasing with time - the fixed length of string is being pulled down). This $V$ is different from $\underline{\mathbf{v}}$, the velocity vector of the mass, given by ...

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 ...

2

Normally you solve an elastic collision with just momentum and energy conservation, because you really don't know what happens at impact. The formulas are given in this answer. For equal masses in one dimension the velocities are exchanged. It turns out your interaction time and acceleration multiply to get the correct velocity change, but since you ...

2

HINT: From conservation of energy, you can find the velocity at the release point; after that it will follow a projectile motion. You can thus write equation of trajectory from the projectile motion.

2

The effect of each $O_2$ molecule on the susceptibility is almost the same in both phases. The magnetic susceptibility of the liquid oxygen is about 1,000 times higher simply because it's a liquid and the density of liquids is about 1,000 times larger than the density of gases (it's more or less true whether you count the density as mass per unit volume or ...

2

The acceleration always points in the same direction as the force. That's because Newton's second law tells us that: $$\vec{F} = m\vec{a}$$ where the force $\vec{F}$ and the acceleration $\vec{a}$ are both vectors. So to work out which direction the acceleration is just ask which direction the force is. In the example you've given consider whether the ...

2

Perhaps an easy way to see where Newton's second law comes from is as follows. Imagine an object sitting in space (so no friction, etc. to worry about). You can push on it (exert some force on it), and see what happens to it. When you push on it with a constant force, you see the object start to accelerate, which means its velocity increases linearly with ...

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