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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 ...


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 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 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

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 ...


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 $$\tag{B} \Delta -(1-z)^2m^2~=~y(y-1)q^2 -2yz q\cdot p~\stackrel{(A)}{=}~y(y-1+z)q^2~=~-xyq^2. $$


2

First assume that $h$ doesn't change very much because you have a large body of water (we can relax this condition later). Let's also assume that the hole is small compared to the depth ($d \ll h$) - we'll relax this too. For this case, the answer is straightforward, you'd use Bernoulli's equations and simply set the static pressure ($\rho g h$) equal to ...


2

The idea of the question is to find the temperature at which the average interparticle spacing is equal to the average de Broglie wavelength. Both of these are averages because the atoms of the ideal gas are not evenly spaced and the velocity (and therefore de Broglie wavelength) of the ideal gas atoms follows the Maxwell-Boltzmann distribution. So this is ...


2

"The size of a single gas particle" is a bad term for what you calculate. A better term would be: "The volume one of the gas particles would occupy, if the total volume were distributed equally among all gas particles" And this translates loosely to "The volume, in which you find one gas particle on average" If you then imagine every particle sitting at ...


2

$$\nabla_{\mu}\left[U,V\right]_{\nu} = \nabla_{\mu}\left(g_{\nu\lambda}\left[U,V\right]^{\lambda}\right)$$ You have two terms inside the parentheses, and you have to apply the derivative to both of them. Myself, I'd just remember that: $$\left[U,V\right]^{a} = U^{b}\nabla_{b}V^{a} - V^{b}\nabla_{b}U^{a}$$, so, we have: $$\begin{align} ...


1

If you note that $$\delta(\cos\theta)=\frac{\delta(\theta-\pi/2)}{\sin\theta}$$ Then you can see that the sine terms actually cancel out.


1

There are a few problems with your derivation. Your main mistake is using formulae that assume the acceleration of the spring is constant (it is not). In one dimensional systems with nonuniform acceleration, we have $$ v(t) = v_0 + \int_{t_0}^t a(t)dt . $$ Furthermore, the $v^2 = v_0^2 + 2a \Delta x$ no longer works for nonuniform accelerations either. ...


1

To be honest, I think it is a superfluous quantity that only arises when integrating in spherical coordinates. I believe the origin of this delta's $sin\theta'$ is contained in $dV' = r'^2sin\theta'dr' d\theta' d\phi'$. As you said, this factor doesn't matter because $$ \int_0^\pi sin^n\theta \delta(\theta - \pi/2)d\theta = 1. $$ Also, suppose we are ...


1

No. First of all, the expression for the velocity that you have there is dimensionally wrong: if $x$ is position, the units of $v$ will turn out to be something$^{metres}$ instead of metres/seconds. So you should stick a constant $C$ in front of your expression so that it takes care of the units, like: $$ v = C2^{2^x} $$ Velocity = derivative of distance ...


1

$ \mathcal{E} = -{{d\Phi_B} \over dt} = -8t$ (Faraday's law of induction) $ E = V/d = \mathcal{E}/d = \frac{-8t}{d} = \frac{-8*3}{15\times 10^{-3}} V/m = -1600 V/m$


1

As Andres Salas said, your kinetic energy should be the negative of the potential energy at the surface. Also, you're using the wrong formula for the potential energy in the crater. $-GMm/r$ is correct only when $r\ge R$.


1

This seems to be an application of the parallel axis theorem. From Wikipedia's http://en.wikipedia.org/wiki/Parallel_axis_theorem: Suppose a body of mass $m$ is made to rotate about an axis $z$ passing through the body's center of mass. The body has a moment of inertia $I_\textrm{cm}$ with respect to this axis. The parallel axis theorem states that ...


1

There are two MR^2 terms added. Once for the shell (the parallel axis theorem requires this additional term, since the torques are being written about an axis passing through the point of contact), and once for the fluid (since it is frictionless, the fluid does not rotate, behaves as point mass, thus only MR^2 suffices).


1

One potential way of going about finding which step you hit is the following. You can parameterize the motion of the particle in the plane by $(x(t), y(t))$, where $t$ is time and $x$ and $y$ are as follows, $$y(t) = y_i + (v_i)_yt - \frac{g}{2}t^2, \quad x(t) = x_i + (v_i)_x t.$$ Fortunately we can kill the parameter $t$, via the substitution $t = \frac{x ...


1

See this page (equation 10, 11 and 12). It will provide the steps needed to show that the equation you have is correct.


1

Here is one tricky way to show that the Dirac representation is invariant under C,P,T-transformations. The free Dirac theory refers to the direct sum $\left( 0 , \frac{1}{2}\right) \oplus \left( \frac{1}{2}, 0 \right)$ of the Lorentz group irreducible spinor representations. As it can be shown, the representation $\left(\frac{n}{2}, \frac{m}{2}\right) \oplus ...


1

Mass conservation assuming stationary conditions: $$\rho(x,t)v(x,t)=\rho(0,t)v(0,t):=\rho_0(t)v_0(t),$$ or equivalently $\rho(x,t)q(x,t)=\rho_0(t)q_0(t)$ since $q=vA$, with $A$ a constant cross section. Therefore $$q(x,t)=\frac{\rho_0(t)q_0(t)}{\rho(x,t)}.$$ Then, we use the initial conditions $$q_0(t)=-\frac{V}{\rho_0(t)}\frac{d \rho_0(t)}{dt},$$ and you ...


1

A sailboat is moving at a constant velocity. Is work being done by a net external force acting on the boat? This is a bit of a trick question it tricks you into thinking there is a net external force. The boat is moving at a constant velocity; that's a given. That means that the net external force on the boat must be zero. But if I use Work = Force ...


1

I am not satisfied with the way @Bernhard answered, since it just shows the maximum velocity, thus only answering partially the question. The air resistance can be written as : $$ R = \frac{1}{2}\,C_x\, \rho\, S\, v^2 $$ Note : The mass of the object is not in this equation. This is very important. Applying Newton's law to one of the object gives at any ...


1

I have used the Darcy Formula together with the following formulas for a quick numeric solution (only a few iterations needed) $$h_f = \frac{\Delta P}{\rho g}$$ $$ f = {\rm Darcy}(Re)$$ $$ h_f = f\,\frac{L}{D}\,\left( \frac{v^2}{2 g} \right) $$ Solve above for $v$ $$ Re = \frac{\rho D\,v}{\mu} $$ Go to step 2 until $f$ converges to a value.


1

When I did the derivation I got this for delta: $\Delta = -y^2q^2 +yq^2 -2zpyq +(1-z)^2m^2$ In the text it says this: $\Delta = -xyq^2+(1-z)^2m^2$ Is there some information missing? I recommend looking up "how to use Feynman Parameters" on google to get more detail, but basically it looks like the derivation is given on this webpage, in example 2. It ...


1

Hint in a parallel circuit, the voltage across each resistor is the same. What is the voltage across the $4 \Omega$ resistor?



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