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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 ...


6

A rough calculation can be done using this frame : On the far left we have Faora and on the far right the soldier she is attacking in the next scene. If we assume the height of a soldier to 1m80 then they are at a distance of 10m It takes her 4 frames to cover that distance in the next scene and the video is 24 fps therefore the speed is $$ V = ...


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.


4

This is just a particle-in-a-box. You only have $\ell, m$ quantum numbers in a more complicated system like an atom--in particular, there needs to be a rotational symmetry and this just isn't that kind of system.


3

In quantum mechanics, an observable is basically an hermitian operator. You can see a definition of it in chapter 4 of Le Bellac's Quantum Physics.


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} ...


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 ...


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) = ...


2

I know you asked for an analytical description, and ray diagrams, but seeing as this is homework related, I'd urge you to try the ray-tracing yourself and I think you'll find the answer without too much trouble. My suggestion would be to try and track how a couple significant rays move through your lens setup. For example, you might try one ray that passes ...


2

If you are looking at visual brightness, then you have to fold the wavelength dependent sensitivity of the human eye to the approximate black body spectrum of the filament into the calculation. At low power the filament will emit mostly infrared radiation, which is not visible. Even at the max. temperature of practical filaments the color temperature of the ...


2

If your action only depends on first time derivatives, it is then not required for the trajectory to have second time derivative -- i.e. an abrupt change in velocity does not by itself give a contribution to the action. In other words, there is no penalty for changing your speed instantaneously. It then means that you can ignore the boundary values for ...


2

$\newcommand{\d}{\;\mathrm{d}}$You should straightforward take the integral that you mentioned: $$\int_{-\infty}^\infty \left| \psi\right|^2 \d x =\int_{-a/4}^{a/4} AA^* \d x + \int_{-\infty}^{-a/4} 0 \d x + \int_{a/4}^\infty 0 \d x =1$$ I won't do the integral and calculate $A$ because it is trivial to do therefore I leave the rest to you.


1

How did you measure perceived brightness? If you used the visible spectrum, the increasing temperature as the power goes up will shift more of the output into the visible. If you measure total output over the whole spectrum, the relationship should be linear. The increasing resistance of the bulb should not be a problem if you truly measure power in. It ...


1

$\newcommand{\d}{\;\mathrm{d}}$ Elaborating on the answer of zeldredge, I want to say why the following expression works: $$P(E_0) = \left| \int_{-\infty}^{\infty} \ \Phi^* \psi \d x \right|^2 \tag{1}$$ Notice that the eigenfunctions of the Hamiltonian spans the space. That is you can write any function $\psi$ as a linear combination of eigenfunctions ...


1

Yes. I'm not sure you can call this "inelastic" collision as that implies that some kinetic energy is lost (due to heat from the collision for instance). If that's wrong feel free to correct me. Start with writing out the conservation of momentum for the situation where two objects merge after colliding. $$p_1+p_2=p_3$$ $$m_1v_1+m_2v_2=m_3v_3$$ You know ...


1

Short Answer: The contact force (normal force if you like) between the pan and the box is 0 because the pan has negligible mass. Long Answer: The key point in this problem is that the pan has neglible mass. Suppose for a second that the pan had some mass $m_p$. After falling a distance of $0.5 \, m$ the box would have velocity $v_b = \sqrt{2gh}$. In the ...


1

Remember that velocity and acceleration are vectors, they have a magnitude and direction. Speed, on the other hand, is a scalar quantity which only has magnitude. If you are moving (in a car) towards the right and apply the breaks, you are decelerating (slowing down)--your acceleration is directed towards the left while your velocity ("speed" in the ...


1

The proof of this identity is entirely analogous to how one would go about proving the Euclidean formula $\vec{a}.\vec{b} = |a||b| \text{cos}(\theta)$. What one has to do is as follows: Find a matrix $M(\theta)$ such that $\vec{b} = M(\theta)\vec{a}$. Use the definition of the inner product $\vec{a}.\vec{b}$. Relate the two equations above to solve for ...


1

I can't comment yet so I'll put this in as an answer. When acceptor and donor dopant concentrations are similar (less than ten times) we simply take the difference as the 'net doping'. This type of semiconductor is called a compensated semiconductor. We can then approximate thermal equilibrium majority carrier concentration at room temperature to the net ...


1

An electret is a insulating material that has been imbued with a semipermanent volume charge distribution by bombarding it with charged particles. I don't know how uniform the charge distribution generally is, but if it were desireable, I'm sure a clever engineer could arrange for an approximately constant volume charge distribution. Beams of charged ...


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 = ...


1

The ping pong ball would lose a tiny amount of kinetic energy to the truck. The truck ends up with a momentum of just under twice what the ping pong ball had. However, energy is 1/2 m*v^2 = 1/2(m*v)^2/m. Since the truck is much more massive than the ping pong ball, it carries much less energy for a given momentum. The end result is that the small amount of ...


1

Balance torques around the corner of the step, so r x cos(38.7) x mg= F x 0.3. F = 114 N


1

What I believe to be true: The bubble changes shape with spherical symmetry, hence the flow is irrotational, and there is a velocity potential. We solve this problem using conservation of mass at the boundary $r=R>a$, and at $r=a$. By conservation of mass: $4\pi r^2*u(r)=4\pi a^2 *\dot{a}$ $\implies u(r)=\frac{a^2 \dot{a}}{r^2}$ Then the velocity ...


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 ...


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

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 ...


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

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 ...


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 ...



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