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0

If we say that the spaceship and earth are very close to each other, let us then imagine that the spaceship, right as it passes by earth, shoots a photon out at $45^{\circ}$ from its direction of travel. The simplest way to attack this problem will be with the transformation of velocities in the x-direction. From our known information, we can determine ...


0

Gauss's Law states that $$ \oint \mathbf{E} \cdot \mathbf{dA} = \frac{Q_{\text{enclosed}}}{\epsilon_0} $$ As you mentioned, since $Q_{\text{enclosed}} = 0$ , $\mathbf{E} \cdot \mathbf{R}$ cannot be everywhere positive or negative since that would make the integral $\oint \mathbf{E} \cdot \mathbf{dA}$ nonzero. Your intuition is correct. In addition, even ...


0

I have to make a couple of assumptions to answer your question. I'm assuming the 4 cm refers to the diameter of said metal rod. I would use Ampere's Law. Picking a circle $C$ centered around the center of the rod with radius $r$ as your closed loop, Ampere's Law states $$ \oint_C \mathbf{B} \cdot \mathbf{d\ell} = \mu_0I_{\text{enclosed}}$$ By symmetry, we ...


0

In doing a force analysis on the system, you can note that the tension in the rope at this point is essentially the force which pulls the far half of the rope. So it may be helpful to consider the force as $T=m_{half \, rope} a$. And we can calculate the mass of part of the rope as: $$ \int_{x_1}^{x_2} \lambda(x) dx = m|_{x_1}^{x_2} $$ The only thing we ...


1

It all depends how tight your coiling is. If the wire doesn't touch itself, i.e. without any contact surface excluded with itself, then the area of the wire exposed is simply the area of the wire : $$ A = 2\pi rL $$ which gives immediately the length of wire you wish (with $d$ the diameter of your wire) : $$ L = \frac{A}{d\pi} $$ $$ L = ...


0

The key idea here is that the volume of the wire before and after flattening it should be almost exactly the same, since you aren't removing or adding any metal, just smushing it around. If $L$ is the length of the wire, $r_{wire}$ is the radius of the wire, $t$ is the thickness of the disk, and $A$ is the desired area of the disk, $$V_{wire} = L ( \pi ...


2

For some function $f$ of $x$, the logarithmic derivative is simply $$ \frac{\mathrm{d}\log f}{\mathrm{d}\log x} = \frac{x}{f} \frac{\mathrm{d}f}{\mathrm{d}x}. $$ You can check that this follows from the chain rule applied to $\mathrm{d}g/\mathrm{d}y$, where $g = \log f$ and $y = \log x$. This is a common thing to see in astrophysics, since if we have a power ...


0

When the pseudoscalar invariant $\vec E \cdot \vec B$ is zero, we have three cases. If $E^2<c^2B^2$ then you can switch to a frame moving with speed $E/B$ in a direction mutually orthogonal to $\vec E$ and $\vec B$ where there is no electric field in the new frame. Solve in the new frame. Then bring it back to the original frame. If $E^2>c^2B^2$ then ...


2

The gravitational force on a small mass $m$ some distance $R$ from the center of a large spherical mass $M$ is given by $$ |F| = \frac{GMm}{R^2}. $$ If your distance from the center is some altitude $r$ above the radius of the Earth's surface $R_\oplus$, the force is $$ |F| = \frac{GMm}{(R_\oplus + r)^2} = \frac{GMm}{R_\oplus^2} \left( 1 + \frac{r}{R_\oplus} ...


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Yes, because gravity decreases with the square of distance. The amount that it decreases is very small, however. Interestingly enough, you can measurably weigh more inn one city than in another, due to differing densities and altitudes.


1

You didn't do anything wrong, just incomplete. As @Sebastian alluded to, the work will be $$W=E_{3R}-E_{2R}=T_{3R}+V_{3R}-T_{2R}-V_{2R}$$ where the subscript denotes orbital position. You already found the $V_{3R}-V_{2R}$ term, now just use the centripetal motion formula $$\frac{mv^2}{r}=\frac{\gamma Mm}{r^2}$$ to find the difference in kinetic energy ...


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The problem is that the equation should be $9m\cos 20= mg \sin 20+\mu mg$ since the friction force is going along the slope.


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I managed to figure this out! It was done using the Clausius Clapeyron equation. $\frac{dP}{dT}=\frac{L}{T(V_2-V_1)}$ By letting $L=\frac{h_f}{m}$ and $V_1=\frac{1}{\rho_1}$,$V_2=\frac{1}{\rho_2}$ gives: $\frac{dP}{dT}=\frac{\frac{h_f}{m}}{T(\frac{1}{\rho_2}-\frac{1}{\rho_1})}$ By solving this differential equation it is seen that: ...


1

The image of each individual optical device forms the (pseudo)object for the next optical device in the light path. You must make adjustments for the position difference of each device, and be sure to keep account of positive and negative values of object/image distances. As a starter for you, the object distance for the concave lens will be be -40 cm.


1

Assume this is on Earth, so you know $g$. Don't try to use any values for $M$ or $k$. Just use symbols. Draw a free-body diagram for the system at rest. From this you can get the relationship between $M$ and $k$ and $g$. You probably also have an expression telling you the relationship between the $T$, $M$, and $k$. Do the algebra and everything should work ...


1

With the help from the comments this now makes sense. \begin{equation} \int \delta(E^2-p^2-m^2)dE \end{equation} With $$E_p^2-p^2-m^2=0$$ Use substitutions \begin{equation} f(E)=E^2-p^2-m^2\quad df=2EdE \end{equation} \begin{equation} \int \delta(f)\frac{df}{2E(f)}=\frac{1}{2E_p} \end{equation} $E(f)$ is easily found by inverting $f$ Thanks!


1

Just a small conceptual hint will do No problem. A hint: Set up Newton's law, $\sum F=ma$. You will see that the sum of all the three forces must equal... yes, what should it equal? I'm confused with the condition at which the block will start sliding What is the difference in the equation mentioned above for a point just after it started moving, ...


1

You're actually dealing with the Potts model, which is a slight generalization of Ising. Not that it really matters, as you won't need any results from Potts. The point of mean field theory is typically to make each site independent of their neighbors, which allows you to evaluate the partition function by only iterating through the possible states of one ...


0

Remember that force is the negative gradient of the potential for a single particle in a potential $\vec F(\vec x)=-\nabla V(\vec x)$. Now for an interaction potential $U(\vec x_1,\vec x_2)$, the forces on each of the two particles are given by the gradients with respect to the coordinate of the particle. One usually denotes that $\vec ...


1

Neutrons have almost the same mass as protons and the unit of atomic mass is (at the moment: see footnote) $\frac{1}{12}^{th}$ of the mass of a carbon 12 nucleus. The mass of nucleons varies a little depending on which nucleus they belong to, but the difference (the binding energy) from their mass when free nucleons is small. So, to a good approximation, the ...


0

When you have a block with a mass of 10 kg, the normal force is approximately 98 N. If the block starts out stationary, it will continue to be stationary until you apply sufficient force to exceed the static friction. At that point it will start to slide, and will continue sliding until the force drops below the force of dynamic friction. For the ...


0

What about the car when there is no force applied to the block? The block will experience no frictional force, so when dealing with static friction, we have to say $f_s \geq \mu_s n$, since there are instances when the frictional force may not act at all or may at at a reduced capacity. So here, $f_s = 18N$, since that is the only force necessary to keep the ...


0

Using the diameter of the muzzle $d$ and the initial distance between the bullet and the end of the muzzle $l$ we can calculate the volume of the gas to be given by: \begin{equation} V\left(l\right)=\frac{\pi d^2\left(L-l\right)}{4} \end{equation} where $L$ is the total length of the muzzle. We therefore find from the ideal gas law that: \begin{equation} ...


-1

Step 1: redraw the network so we can more clearly visualize the series/parallel relationships: *------*-------*-------* | | | | | 3ohm 6ohm | | | | | 6V *---*---* 3ohm | | | | 4ohm | | | | *----------*-----------* Step 2: note that there are ...


-2

Note that 3 Ω on the left and 6Ω on the top are in parallel: 6 Ω (on the top) // 3 Ω (the one on the left) = 2 Ω Now, we replace 6 Ω//3 Ω with 2 Ω found. Then we have 2 Ω in series with 4 Ω (in the middle): 2 Ω + 4 Ω = 6 Ω Now, in the middle we have only 6 Ω and 3 Ω in parallel on the right: 6 Ω // 3 Ω (the one on the right) = 2 Ω 2 Ω is the equivalent ...


1

$I_0$ is the intensity of light before it hits a polariser the original intensity of the beam, so called. You need it because you need to compare it to the intensity after it exits the polariser so that you can calculate your fraction of incident intensity. this fraction requested by the problem is $I\over I_0$, but $I$ refers to intensity of light exiting ...


1

Clearly $a$ has the same dimension of $x$ (see the argument of root or of $\sin^{-1}$) so the left member is dimensionless (ratio between dimension of x: remember that differential dx count in dimensional calculus!), and the second member too has to be dimensionless: so n=0.


1

Integration is finding the area under a curve that isn't necessarily straight. If you have a velocity time graph and find the area under it, this gives you the distance travailed. If you have a acceleration-time graph the area under it is the change in velocity. There are several techniques to integration, which I will not go into here. As mentioned in the ...


1

Assume plane waves. The tangential boundary conditions show that the transverse electric and magnetic field vectors must stay in the same direction on transmission or reflexion from the interfaces, assumed aligned with the wavefronts. Since we know the direction of the waves, let's say the $\vec{E}$ fields are all in the $\hat{X}$ direction, the magnetic ...


0

Well, I've done some calculus to your problem. The problem, in fact, is that $|\vec{r_i}-\vec{r_j}|$ everywhere. Also the temporal derivative of that is pain. Of course you can write it without that angle you commented if you write the derivatives in terms of escalar products. However, I'm not going to follow that. You have a system of 3 bodies, isolated, ...


0

Okay, I got the solution myself. The integral was causing problems, see my question on Math SE for the answer. The solution is $ \Delta \varphi = \frac{\pi R w_0}{v} \sqrt{\frac{2M}{2M+5m}}$


0

Because it is rotating with constant w0 you should not use conservation of angular momentum and for the same reason: The mass $m$ and $M$ are irrelevant to the problem. May be the author intended to say differently: INITIALLY a sphere of mass M is ... with constant w0 the problem is much simpler. Initially ... Yours response appears to indicate ...


1

Can you tell from the image below if Q1 and Q2 are attracted or repelled? No, you do not have enough information. Will Q2 only be attracted to the sphere if Q2 is enough bigger than Q1? For any nonzero values of Q1 and Q2 you can compute the distance at which there is no net force. Will the positive charge inside the shell attract electrons interior to ...


1

Initial kinetic energy is $K_1=\frac{1}{2} m (v_x^2+v_y^2)$ with potential energy $U_1=0$. At the apogee, the potential energy is $U_2=m g h$ and the kenetic energy is $K_2=\frac{1}{2}m v_x^2$. Equate the two sums to get your answer. $$U_1+P_1 = U_2 + P_2 $$ $$0+\frac{1}{2} m (v_x^2 + v_y^2) = m g h + \frac{1}{2} m v_x^2 $$ $$ \frac{1}{2} m v_y^2 = m g ...


-2

Where U is the intial velocity,g is the gravitational constant,R is the range of the projectile and ∅ is the launch angle between the projectile and the flat ground (imagine the flat ground as the x-axis with the projectile inclined at an angle ∅ on it).


4

Yes, because acceleration also includes change in direction. For example, a race car on the track goes in a circle. If its speed is 150 mph for the entire race, it is still accelerating because it is not going in a straight line.


0

We can use the following the equation to solve for $B$. $$E = cB$$


2

1) On integrating dt on the RHS we get a +c(constant of integration) but why is there no +c on the LHS while integrating dv? If we start with: $$ dv = adt $$ and integrate both sides then we can indeed have a constant of integration on both sides: $$ v + C_1 = at + C_2 $$ but we can just subtract $C_1$ from both sides to get: $$ v = at + (C_2 - C_1) = ...


0

In the first step ( $\int dv=a\int dt$), we get $v+c_1=a(t+c_2)$. and therefore $v=at+(ac_2-c_1)$. The term $ac_2-c_1$ is constant (because $a$ ,here, is constant). Instead of tediously writing the integration constants for every step, we can write one at the ending, since the last constant ($c$, in this case) absorbs all the other constants which have ...


0

That discussion is about a spherical shell and does not apply to the Earth nor to the sun. In the latter situations assuming uniform mass distribution we know that time is dilated by $t(\infty)/t(r)=\sqrt{g_{00}}=\sqrt{1 - \frac{2GM}{rc^2}}$, where $r$ is the distance from the center. For a spherical SHELL however we have to go back and forth between the GR ...


1

You are right in stating that potential and hence potential differences are dependent on field. The relation in fact is $\mathbf{E} = -\nabla V$ Hence, as we can see, if $E$ = 0, then $\nabla V$ is in fact constant, not $V$. Now, to compute the potential, we can rely on coloumb's formula, taking $V$ at infinity t be zero, for a differential ...


1

Definition of potential difference is the amount of work per unit charge to move a charged particle from one place to the another place. The potential difference between point $a$ and point $b$ is as below, $$ V_a - V_b = - \int_{\mathbf{r}_b}^{\mathbf{r}_a} \mathbf{E}\cdot \mathrm{d}\mathbf{r}.$$ What we call as potential with $V=\frac{kQ}{r}$ is the amount ...


0

Marshmallow is a Solid-Gas Colloid named Solid foam, consisted of a Dispersed phase and Dispersed medium.


6

Marshmallow is traditionally made by stirring a hot supersaturated solution of sugar and gelatine or agar into whipped egg whites. On cooling a material with (at least) four phases present is formed. The phases are: the protein in the egg and gelatine (or polysaccharide in the agar) form an elastic solid held together by crosslinks between the protein ...


0

The electric field due to the charge is uniformly distributed in all directions in the space.., the flux will also be uniform for the charge on all directions regardless of shape if we consider the surface vector is parallel to the electric field lines.


0

Well, the simple point is Kinetic energy depends upon the mass and the square of velocity. Even if the mass of the object in this case molecule is relatively greater than that of the other molecule, as long as the velocity is greater enough to balance the average kinetic energy, it will be uniform. To put it simply K.E = 1/2*(m*v^2).


0

According to Gauss's law, the flux through a closed surface is equal to $q_{enclosed}/\epsilon_o$. This doesn't talk about the type of closed surface that encloses the charge. Hence, the flux remains the same regardless of the shape or size of the surface in question. (as long as the charge remains enclosed in it.)


0

Kinetic energy is related to both the mass and velocity of a particle through the equation $T= \frac{1}{2}mv^2$ where $T$ is the kinetic energy of the particle. So you are correct that the oxygen particles will likely move slower, because they are more massive, but as long as the product of the mass and the velocity squared of the oxygen particles (on ...


1

You should read this page :http://en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle The Huygens principle explain intuitively why the wave will spread after being "cutoff" by an obstacle, as the spherical sources at the edge will not interfere anymore with the adjacent ones (those being stopped by the obstacle) to form a plane wave. As shown in the ...


1

One approach: the liquid-drop model says they should have the same $R\propto A^{1/3}$, but different Coulomb energies proportional to $Q^2/R = Z^2e^2/R$. To wit: the self-energy of a uniformly-charged sphere with radius $R$ and charge $Q=Ze$ is $$ U = \frac35 \cdot \frac1{4\pi\epsilon_0} \frac{Q^2}{R} = \frac35 \cdot \alpha\hbar c \frac{Z^2}{R}. $$ ...



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