# Tag Info

43

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 ...

19

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 ...

11

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, ...

6

The period of an elliptical orbit is given by: $$T = 2\pi\sqrt{\frac{a^3}{GM}}$$ where $a$ is the semi-major axis. For a circular orbit of radius $r$ we have $a = r$. The two orbits you show do not have the same semi-major axis, so they do not have the same period. However if the elliptical orbit had $a^3 = 4r^3$ then the period of the elliptical orbit ...

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

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 $$\begin{pmatrix} \ddot{x} \\ \ddot{y} \end{pmatrix} = \begin{pmatrix} 0 \\ -g \end{pmatrix} - \beta ... 4 Following the same orbit at greater velocity would mean that the satellite has greater acceleration (its velocity is changing at a greater rate than if it follows the same route more slowly). But the only force acting on it is earth's gravity, which creates a particular acceleration at any given altitude above the earth. It can't go any faster or slower ... 4 The wind is certainly doing work, because it applies a force and the point where the force is applied is displaced. However it isn't doing any work on the boat, it's doing the work on the water. The key point is that the net force on the boat is zero. We know the net force on the boat is zero because the boat is moving at constant velocity - if the net ... 4 The distance travelled in a time t is:$$ s = ut + \tfrac{1}{2}at^2 $$So the distance travelled between t and t - \Delta t is:$$\begin{align} \Delta s &= s(t) - s(t - \Delta t) \\ &= ut + \tfrac{1}{2}at^2 - u(t - \Delta t) - \tfrac{1}{2}a(t - \Delta t)^2 \\ &= u\Delta t + \tfrac{1}{2}a(2t\Delta t - \Delta t^2) ...

4

The faster you go, the less velocity you theoretically can gain from a gravity assist. The reason for this is that the faster you go the harder it is to bend the orbit. To proof this we have to use the patched conics approximation, which means that while within a sphere Kepler orbits can be used. The sphere can be simplified to be infinitely big, since the ...

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

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

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 $$\delta \mathcal L = - (\delta ... 3 You know that in GR you need a locally Minkioski spacetime. This, in each point of your manifold you can change the coordinates so that the metric is diagonal, and the square of the infinitesimal displacement is ds^2=\left(ct\right)^2-x^2-y^2-z^2. So here is where the c come from. Then, when you want to compute the coupling constant k=\frac{8\pi ... 3 Consider this diagram showing the three columns you describe all connected to the same body of water: Your question asks whether the three pressures P_1, P_2 and P_3 will be the same. The answer is obviously yes, because the columns are all connected to the same body of water. For example if P_1 > P_2 then water would flow from the base of ... 3 It is an axiom in most formulations of QM that the state describing te system is a vector on Hilbet's space. So do not swet about it unless you wanna change the axioms (which will not be a bad idea) 3 Hint: Recall that$$\tag{A} q\cdot (q+2p) ~=~(p^{\prime}-p)\cdot (p^{\prime}+p) ~=~p^{\prime 2}-p^2~=~m^2-m^2~=~0 .$$So$$\tag{B} \Delta -(1-z)^2m^2~=~y(y-1)q^2 -2yz q\cdot p~\stackrel{(A)}{=}~y(y-1+z)q^2~=~-xyq^2. $$3 One can get an order of magnitude estimate of the maximum speed attainable by gravitational slingshots without doing any real calculation. The 'rough physics' reasoning goes as follows: The gravitational field of the planets used for slingshots needs to be strong enough to "grab" the speeding spaceship. As a planet cannot "grab" a spaceships moving faster ... 2 The proof is most easiest if we use the vector notation. We have$$\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

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 ...

2

$(4)$ is false. Note that by the same logic, $v = \frac{\mathrm{d}|\vec{x}|}{\mathrm{d}t} = \partial_tR = 0$. Please note that you should ask about a specific physics concept and not just throw "Here's a calculation, show me where I'm wrong" at us.

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

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 ...

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