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0

$ \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

its very simple, follow these steps 1)as the circuit tends to infinity, consider the equivalent resistance to be R(e) 2)consider the parallel resistors i.e., BC and the rest to the right of BC. these are in series with AB 3)now the total resistance to the right of BC will be equal to the effective resistance of the total circuit R(e), as the circuit tends ...


-2

We know that $$\sin (\theta) = y/x$$ is true, where $y$ is the magnitude of the first force and $x$ is the magnitude of the second. Look at a unit circle! It is pretty easy to see that if $y$ is halved so that $y/x = 0.5$, then the original angle cannot be $90$, otherwise, $\sin45$ would have to be equal to $0.5$, but it is equal to $\sqrt{2}/2$.


0

Since your particle should follow a parabola, all you need is the formula for the area under a parabola, which is $A = \frac{2}{3} R\cdot H$ A derivation of this can be found here http://www.math.nmsu.edu/~breakingaway/ebookofcalculus/HandsOnUnitsInvolvingIntegrals/AreaUnderParabola/AreaUnderParabolaWebPage01.html. Using differential calculus and ...


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 $r\ge R$.


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 ...


0

Certainly late for your homework problem set, but for future readers: I think it best to go to the source, Ernest Rutherford, "The Scattering of $\alpha$ and $\beta$ Particles by Matter and the Structure of the Atom", London, Edinburgh and Dublin Philosophical Magazine and Journal of Science, Volume 21, Issue 125, pages 669-688 (1911). This is the paper in ...


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 ...


0

I get the same answer as others but in a different way. First I look at the slip speed $v_s(t) = \omega(t) r + v(t)$ and find the time needed to get $v_s(t)=0$. The time functions of the motion depend on the constant friction (until rolling starts) with equations $$ v(t) = v_0 - \mu g t \\ \omega(t) = \omega_0 - \frac{\mu m g r }{I} t $$ With the general ...


0

Your expressions are all correct, except for your work due to torque. Because the cylinder isn't rolling, $\theta \neq \frac{d}{R}$. Torque is constant though, so we can write $\theta = \omega_0 t -\frac{1}{2}\alpha t^2$. Furthermore, the work due to torque is negative: $W = \int F \cdot ds + \int \tau \cdot d\theta = \mu_k mg d - \mu_kmgR\theta$ And then ...


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 ...


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.


0

(In the laboratory frame of reference) Using the principle of conservation of linear momentum (i.e. momentum before collision = momentum after collision) we know that: $$m_1v+m_2(0)=m_1v_1+m_2v_2$$ since both masses are the same we get: $$v=v_1+v_2$$ now using the principle of conservation of energy (i.e. intial energy = final energy) ...


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) ...


0

OK, i have since edited this, but the results are still off. What am i missing? My first step is just to be able to calculate the x,y values for the given angle. I have omitted irrelevant methods in code, i just need someone to assist with the PHYSICS portion. i a just missing something. CODE BEGIN const float g = 32.2; // gravity 9.8m/sec^2 or ...


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 ...


0

The first thing I should point out is W$\cdot$x. x is the displacement. Direction is important. That's really not the most important thing for your answer though. The punch line, Work is done if there's a force (hence change in velocity) and it causes displacement So, if something's moving a constant velocity, it doesn't have any force acting on it ...


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 ...


0

Good one. The net forward thrust afforded by power of the engine, wind, rowing etc., is constantly overcoming and exceeding the retarding force of resistance by water, air, etc. So, the movement of boat in water and air is constantly impeded or constrained by water and air resistance-tending to bring the boat to halt. The forward thrust is constantly ...


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 ...


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 ...


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


0

The distance is a function of the angle. I am pretty sure you won't be able to obtain the inverse function analytically, not taking into account air resistance, so you should solve the relevant equation numerically, using one of the well-known methods, such as dichotomy or the secant method.


0

A couple things, first you are not discussing air resistance correctly. The drag depends on the current velocity, which is a dynamical quantity, not just on the muzzle velocity. You need to use the current velocity at any step of the calculation. Second, in broad terms, you can think of the problem you face as one of root finding. You have some function ...


0

The NCEES: FE Reference Handbook has some good material on fluid flow through a submerged orifice in its fluid mechanics section. You can search for it online. NCEES will provide you with one free of charge.


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.


0

From a Wolfram article we get the simplified Bernoulli equation: $$Q = a c \sqrt{2 g h}$$ Where $Q$: the flow rate ($\mathrm{m^3/s}$) $a$: the area of the hole ($\mathrm{m^2}$) $c$: flow coefficient (dimensionless) $g$: the gravity acceleration ($\mathrm{m/s^2}$) $h$: the depth of the hole ($\mathrm{m}$) That is valid for a small enough hole, but since ...


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 ...


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?


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 ...


0

If you want to take the sum of moments, you have to take **all the external forces ** into account. The summation of moments has no preference for a certain force, allowing it to be in the equation. All forces are in the summation, only their resulting moment can be zero. If we look at the following example, assuming that all the spacings, e.g. ...


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, ...


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 ...


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.


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 ...


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, ...


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 ...


1

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: ...



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