# Tag Info

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 ...

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.

3

Use kirchhoff's first law, so for two resistors in parallel: $$I_\text{total}=I_1+I_2$$ Then just use I=V/R $$\\\frac{V}{R_\text{total}}=\frac{V}{R_1}+\frac{V}{R_2}$$ The voltage across any component, whether it be across resistor 1, 2, or the whole parallel portion of the circuit, is the same. It just cancels out so you can divide both sides by V.

2

This is more intuitive if you think in terms of conductance, which is the inverse of resistance (1/R). When you put two equal resistors in parallel, you double the overall conductance. Why? You are adding a second path for current to flow, so you double that flow. Unfortunately, we tend to speak mostly in terms of resistance, which makes the math a bit ...

2

Assuming that you mean $R_3$ is the initial distance from one object to the other, and given that gravitational force goes as the inverse square of the distance $F \propto \frac{1}{R^2}$, it follows that if $R$ increases by $3R$ (making the final distance $(3+1)R = 4R$), then the force changes by $\frac{1}{4^2} = \frac{1}{16}$ - it does indeed become 16x ...

2

Yes, you can use the method of images because uniqueness of the solution is guaranteed when you Know the total charge of an equipotential surface without knowing the value of the potential itself I'll summarize a procedure to obtain the correct answer: Application of Gauss law tells us that there must be total charge -q on the inner surface then because of ...

2

Firstly, $m$ does not have to be an integer, it is entirely possible for $m$ to be 1/2 for instance. Your points ,1-3 are fine. There are is a maximal and a minimal value of $m$. Call the maximal value $M$ (we have to call it something). Now we can apply the lower operator any number of times, each time it lowers the value of $m$ by a full integer ...

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: $$v = at + (C_2 - C_1) = ... 1 Just write out the product$$u_\alpha u^\alpha = g_{\alpha\beta}u^\alpha u^\beta= g_{00}u^0 u^0\ ,$$since u^i=0\ ,\ i=\{1,2,3\}. Then imposing$$u_\alpha u^\alpha = -1 \quad \Rightarrow \quad g_{00}=-1 \ .$$I think the rest you can do on your own :) 1 Initial kinetic energy is K_1=\frac{1}{2} m (v_x^2+v_y^2) with potential energy U_1=0. At the apogee, the potential energy is U_2=m g h and the kenetic energy is K_2=\frac{1}{2}m v_x^2. Equate the two sums to get your answer.$$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 ...

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

$I_0$ 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

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 ...

1

Since the charge Q have to be the same for both capacitors and you need more voltage to to push that charge in the capacitor with less capacitance then you must have more valtage difference in $C_2$ The mechanical analogy is a configuration with 2 springs in parallel that move the same distance from their equilibrium position need more force on the spring ...

1

The working voltage of a capacitor depends on the dielectric strength of the insulator. While electrical breakdown is actually a very complicated process with lots of non-linearities, you can simplify the design of a capacitor by saying "the electric field on the insulator must not exceed X". Once you have said that, and you realize that the electric ...

1

I think your problem is that you're adding the electric fields like scalars, rather than breaking then into their vector components. You have: $E_1 = \frac{q}{2\pi\varepsilon_0\rho_1}$ and $E_2 = \frac{-q}{2\pi\varepsilon_0\rho_2}$ What you should have is $E_1 = \frac{q}{2\pi\varepsilon_0}(\frac{1}{\rho_{1x}}\hat{x} + \frac{1}{\rho_{1y}}\hat{y})$ and ...

1

Since this is a homework problem, I won't provide a full solution, but here's a nudge in the right direction. Take a look at these two plots of the effective potential: k = -1, $\alpha$ = 1, L = 0.25 k = -1, $\alpha$ = 1, L = 1 What's different about these two effective potentials? We only changed $L$ between the two graphs; what does that imply about ...

1

This section might help: http://en.wikipedia.org/wiki/Lightning_rod#Should_a_lightning_rod_have_a_point.3F Which also states that : Finding that moderately rounded or blunt-tipped lightning rods act as marginally better strike receptors.

1

Here we are talking about instantaneous velocity. So,its -20 m/s. And the velocity will be always tangent to the circular track. So,it will be 20 m/s in magnitude every second. But the direction will be different so different values in x & y - axes

1

As the particle travels around the cylinder (having radius r) and the taught string wraps around the cylinder, the path of the particle will trace out a spiral with radius that decreases as a function of the cylinder radius. In other words, the radius of the spiral (let's call S) equals the string length at any given point. All you need to do is find the ...

1

Consider these two arrangements of charges: Suppose we ask what is the flux through the surface $S$. If you look at figure (a) with two positive charges the flux lines from the two charges travel in opposing directions and will cancel each other out at $S$. So the flux through $S$ will be the flux from one charge minus the flux from the other charge. ...

1

You set $\rho$ equal to one for no reason. In detail the expression for $\Gamma$ is: $$\Gamma^1_{01}=\frac{1}{2}\sum_\rho g^{1\rho}\left[\frac{\partial g_{1\rho}}{\partial x^0} + \frac{\partial g_{\rho 0}}{\partial x^1} - \frac{\partial g_{01}}{\partial x^\rho}\right].$$ And since $g_{01}$ equals zero (since your metric is diagonal), all four partial ...

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

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

Two identical hoses with the same pressure difference between their ends will carry twice as much water as each does individually. So you get twice as much flow (current) for the same pressure difference (voltage), which is another intuitive way of thinking about Dave's Answer that it is more enlightening in this case to think in terms of conductances rather ...

1

Answer if anyone is interested. In the end the areas outside the inner edges of the wire cancelled by symmetry and so the surface i was looking for was the area enclosed.

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