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This addresses the broader question of potential energy minima, not the specific situation of a conductor/magnet/current. You often see statements that lower potential energy states are preferred in some way. Personally I think there are a few issues with such statements, especially for systems in which only conservative forces act. Nonetheless, I think ...


1

By symmetry, you need only consider the force on one of the charges on the corner, say the top right, due to the other four. There will be a repulsion from the other corners pushing it away from the centre, so you must pick Q to balance this with an attractive force of just the right size. One way to work this out is to calculate the vector forces ...


0

Find the torque on the rod (around the axis of rotation) and remember that: $$torque = \frac{dL}{dt}$$ where $L$ is angular momentum of the rod. Then use the relation between $L$ and angular velocity, remembering: which quantities are constant in time what is the rate of change of angular velocity.


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In this case, a variant of Newton's 2nd Law adapted to rotational motion can be used: $$T = J\alpha$$ Where $T$ is the applied torque, $J$ is the polar moment of inertia, and $\alpha$ is the angular acceleration. Compare with: $$F = ma$$ The motion we are interested in occurs about the hinge, so let's consider all our axes to be about the hinge. The ...


0

I found the answer to be .756N by multiplying .63N and sin(36.87).


0

Here is a big hint... I have to believe you can finish it from here. The key is "similar triangles".


6

There are two parts to your question, and I'll address them individually, assuming that the room+ball system is an isolated system, i.e. no work is being done on the system, and no energy is being added / removed. Part I: How cold will it get? This depends on the size of the room, the materials that everything in the room is made of, the size and material ...


1

To remember better how it works, you can just tell yourself that units follow the same operations as numbers. You obviously knew you had to do a product to get "30", now do the same with units. (as Danu is implying)


2

You should think about the definition of torque: $$ \vec \tau = \vec{r}\times \vec F\implies \tau =rF\sin\theta$$ where $\theta$ is the angle between $\vec r$ and $\vec F$, the second equation refers to the magnitude of all quantities. If the units of $F$ are Newtons, and the units of $r$ are meters, then what should the units of $\tau$ be? Remember that ...


2

You correctly identify the residual velocity of the water after bouncing off the bucket as a critical parameter in the calculation. Where you go wrong is in assuming that you can assign any value you want to it. If your bucket's bottom was shaped in such a way as to "turn around" the water jet hitting it, then you would have the maximum possible momentum ...


2

The "missing" kinetic energy is still there ... it's now in the rotation of the yo-yo.


2

Remember how work is defined. The key word is displacement. I think that when you wrote $Fd$, you considered $d$ to define a single point, and not displacement. Now back to your problem. The work done by a net force is $$W=Fd=mad=ma{(}\frac{v_{f}^{2}-v_{i}^{2}}{2a})=\frac{mv_{f}^{2}}{2}-\frac{mv_{i}^{2}}{2}=\Delta E_{c}$$ The single diference is that I ...


0

OP's identity generalizes to the truncated BCH formula $$\tag{1} e^{\hat{A}}e^{\hat{B}}~=~e^{\hat{A}+\hat{B}+\frac{1}{2}\hat{C}} $$ where the commutator $$\tag{2} \hat{C}~:=~[\hat{A},\hat{B}]$$ is assumed to commute with both $\hat{A}$ and $\hat{B}$, $$\tag{3} [\hat{A},\hat{C}]~=~0\quad \text{and}\quad [\hat{B},\hat{C}]~=~0. $$ [In particular, the ...


0

For visibility I will post this as an answer as well. The problem in my text book wanted me to fit the numerical/dimensionless approximation to the graph in the text book, to estimate at what value of $T$ the graph would reach half its maximum value (in Figure 1.14 in Schroeder's Introduction to Thermal Physics that's $C_V=3R$). The estimation of ...


1

take only first 2 resistors and rest as $x$ now the vertical resistor and your $x$ will be in parallel, effective resistance would be $Rx/R+x$ with series in horizontal resistor. Now equivalent resistance would be $$Req. = R+ (Rx/R+x). $$ take Req. as $x$ again Form quadratic equation and solve for $x$. This will be the answer.


1

In this context, the least ambiguous reference frame is the comoving rest frame. In essence, this is the local frame moving along with the local Hubble expansion. We can accurately determine earth's velocity with respect to this comoving frame (and thereby obtain our so-called peculiar velolocity) by subtracting out the dipole anisotropy from the Doppler ...


0

You're asking the spring to remove all (or almost all) of the kinetic energy of the incoming object. That's the same as asking what kind of spring can store that much energy in a compression of 1m. If the spring is light enough, you can ignore its mass and accelerations. As it gets more massive, then you don't need as much of it to decelerate, because ...


1

The answers already in here are good; unfortunately the integrals that arise are quite nasty, and don't have solutions in terms of elementary functions. Here is some more detail, in the special case when the tubes have zero length (so they are just charged circular loops), and further they have the same radius $b$, with separation $d$. You'll see that this ...


0

You are given the deflection. You can use that to compute the force-it is $200 \cdot 0.5+50 \cdot 0.5^2+10\cdot 0.5^3=113.75$ lb. The equivalent linear spring constant is $\left. \frac {dF}{dx}\right|_{x=0.5},$ but I get $257.5$ lb/in for that, close to the book but not exact. The other thing would be to draw a straight line through $(0,0)$ and $(0.5, ...


0

It looks like in this problem the rain moves at whatever $v$ the train is currently at, giving an infinite amount of energy as $lim_{t\rightarrow \infty}$. However it doesn't look like your using energy anyway, just be careful if you do. As BMS said, use the product rule. This gives you $F= {\Phi}{v(t)}+\frac{dv}{dt}{(m+\Phi)}$ then subtract $\Phi v(t)$ to ...


0

See this for variable acceleration problems. Lastly, the acceleration is a function of velocity $a(v)$. Then the time as a function of velocity us $$ t(v) = \int \frac{1}{a(v)}\,{\rm d}v $$ and the position as a function of velocity is $$ x(v) = \int \frac{v}{a(v)}\,{\rm d}v $$ which need to be back-solved for $x(v(t))$


1

$(1.000 miles / 1 hour)$ * $(1609 meters / 1 miles)$ * $ *(1hour/ 3600 seconds)$ Thus $.4469$ meter/second$


0

Keeping in mind you know basic calculus: (took the image from here) You have a expression for the acceleration where $$acceleration, \left(\frac{dv}{dt}\right)=-av^2$$ I put a minus as the retardation was mentioned I guess you might have learned integration by separating the variables. After integration, you will get and expression for velocity. As seen ...


0

Correct me if im wrong The circuit can be reduced to Current should be I = V/R = 4/3V The voltage drop across the 6 resistor is 6/9 * 12 = 8 and the voltage drop over the 3 should be 12-8 = 4 Then we look at the circuit again The Voltage drop across each 'section' of parallel resistors is the same. So V across the 8 and 4 is 4. To calculate Vo ...


0

You need to solve the differential equation of motion. In this case, that equation is given by Newton's second law. (That's a hint. Ask for clarification here if you need to.)


1

You should be careful, since you need to take account of the force that the rain exerts on the trolley, or the momentum of the rain. Your second approach does this rather nicely, (with the assumption that the rain falls vertically, and hence doesn't contribute to the initial momentum). In the first approach, you could redo it to add the force that the ...


2

Newton's 2nd law in differential form (ignoring vectors) is $$F_\text{net}=\frac{dp}{dt}=\frac{d}{dt}\left(p_\text{train} + p_\text{water in trolley} + p_\text{rain just hitting trolley}\right). \tag{1}$$ You must take into account the change in momentum of the rain that occurs when it falls into the trolley and accelerates up to the speed of the train. ...


0

Many thoughts later, I think the correct free body diagrams are those : So, the vector equations are : Block : $$ m_2\vec{a_2}=\vec{W}_{weight-of-block}+\vec{F}_{plan-acting-on-block}+\vec{F}_{friction-from-plan-to-block}+\vec{N}_{normal-from-plan-to-block} $$ Plane : $$ ...


1

For simple harmonic motion, the restoring force is proportional to the displacement. So if you can compute the force $F$ on the drum when you displace it by a small distance $d$, then your "spring constant" $k = \frac{F}{x}$. I am sure you can compute that force (from the weight of the displaced fluid - Archimedes' principle). But warning - before you can ...


2

This looks like a homework problem, so I'll just give you a hint, rather than give you the whole answer. The hint is that according to Archimedes' principle, the buoyant force on a body that is fully or partially submerged in a fluid is equal to the weight of the displaced fluid. So the net force on the barrel at a given time is the weight of the amount of ...


0

It is worth noting that the force due to air resistance is usually modeled as $$\vec{F} \propto -\vec{v} \ \ \ \text{or} \ \ \ \vec{F} \propto -|v|^2 \hat{v} $$ Intuitively this makes sense: the faster you go, the more drag you should experience. The minus sign means that the force acts in opposition to the direction you are travelling. Since the force is ...


2

A telescope with two convex (converging) lenses is a Keplerian telescope. The lens with the longer focal length is the objective, and the shorter focal length lens is the eyepiece. Since it is explicitly stated that the lenses are thin, you can use the thin lens equations: $$ \frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f} $$ where $d_i$ is the distance to the ...


-1

L(angular momentum) = (moment of Inertia) x (angular velocity)


7

The diagram is misleading. Look at this: At any moment in time, you have the following forces on the particle: Gravity Tension in the string When you are at the bottom of the path, the tension in the string is equal to the tension needed to counter gravity, PLUS the tension needed to keep the mass in its path (in other words, to keep the string ...


5

Please note that in the picture, there are two forces acting: 1) the weight, mg, which acts vertically downward, and does not change, and 2) the tension in the string, Z, which points from the mass to the point the string connects to the ceiling, provided the string remains taut. Z varies with time periodically. These two forces combine to give the ...


6

Consider this for the upper ball: The angle of the inclined plane is simply derived from geometrical considerations. Looks pretty easy to solve, right? Once you know the normal force of the inclined slope, reflect and apply it to the bottom ball to complete the problem.


1

I'm not sure how one can know that the half maximum corresponds to $kT/\epsilon \approx 1/3$ without resorting to the formula for the heat capacity. Still, notice that there are only two energy scales, i.e., $\epsilon$ and $kT$, in the problem. Then, whatever (dimensionless) number that determines whether the equipartition holds or fails has to be the ratio ...


1

The magnetic force that one wire exerts on another will be uniformly distributed along the length, given the wires are parallel. Also, assuming the wires are thin, no considerable torque will be applied to cause rotation of the wire about it's axis. Even if that torque were considerable, it would only cause the wire to spin about its axis without any ...


2

The heat capacity of an Einstein solid is given by \begin{equation} C = Nk \left(\frac{\epsilon}{kT}\right)^{2} \frac{e^{\epsilon/kT}}{(e^{\epsilon/kT}-1)^{2}}, \end{equation} where $N$ is the number of degrees of freedom. So the value of the energy quantum $\epsilon$, or more precisely the ratio $x\equiv\epsilon/kT$ matters! The above equation tends to the ...


0

Are you talking about the forces two parallel current-carrying wires exert on one another? Given two current-carrying wires, $a$, and $b$, we can determine the force exerted on wire $b$ by wire $a$ with $$F=(µ_oI_a/2πr) I_bL$$ Where F is the force exerted, $µ_o$ is the magnetic permeability of a vacuum, $I_a$ is the current flowing through wire $a$, $I_b$ ...


0

You ended up multiplying by 1. $9.06\cdot 10^8 km \times \frac {1000m}{1km}\times \frac {1000mm}{1m}$ $\ \ \ \ \ =9.06\cdot 10^{14}$


1

$$9.06 \times 10^8 km \times \frac{number\, of\, mm\, in\, 1 km}{1 km} = 9.06\times 10^8 km \times \frac{10^6 mm}{1 km} = 9.06\times 10^{14}mm$$


2

You really, really have to be able to draw a diagram to understand many physics problems. Here is my interpretation of what you write above (with apologies for poor lining up of various lines due to the limitations of the drawing package I had to hand). The normal force acts at right angles to the surface and is labelled as $N$. The weight of the ...


3

If $V = \frac{m}{\rho}$, $dV = \frac{m}{d\rho}$ This is the source of your error. You can re-write the above as $\rho V = m$, and this yields $\rho dV + Vd\rho = 0$, or $V \frac d {dV} = -\rho\frac d {d\rho}\,$ as a differential operator. This leads directly to the alternate form for the bulk modulus $B = \rho \frac{dP}{d\rho}$. Being a bit more ...


3

You have to use the differentials properly: If $V = \frac{m}{\rho}$, then $$ \mathrm{d}V = \frac{\partial V}{\partial \rho}\mathrm{d}\rho = -\frac{m}{\rho^2}\mathrm{d}\rho$$


0

Yes you would actualy see the car with its speed added with your speed.


3

As Emilio points out the gravitational attraction by pluto is on the order of $10^{-14}\textrm{ m s}^{-2}$. However this is not an acceleration that can be felt in any meaningful way - In a flat gravitational field movement due to acceleration by gravity is identical to inertial movement. However as it happens the gravitational field from pluto is not ...


3

You have obtained several interesting answers. Here is the one I prefer, which actually is a variation of gj255's answer. Let us attack the problem by computing the function $$f(\mu) := e^{\mu X} Y e^{-\mu X}\:.$$ This function verifies $$f'(\mu) = e^{\mu X} X Y e^{-\mu X} - e^{\mu X} YX e^{-\mu X} = e^{\mu X} [X, Y] e^{-\mu X} = \lambda I\:.$$ ...


1

Your understanding is correct. And switching between different perspectives (what physicists would call different inertial frames of reference) like that is a very useful tool in physics, because it turns out that the laws of physics have the same form no matter which inertial frame of reference a problem is described in. For more information, see ...


3

Rough impressions can be misleading. The other car really is moving 200 km/h from your point of view. One thing to keep in mind is that you tend to perceive motion more readily when it is closer to you. This is at least partly due to the fact that you really only can see angular speed across your field of view (like degrees per second). To convert this into ...



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