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8

I doubt if your textbook makes it explicit, but the only sacred tenet in here is to respect dimensional homogeneity. One can make no sense of the sum of quantities with different dimensions.


7

From the commentary to the question, the textbook in question appears to be a mathematics textbook rather than a physics textbook. In mathematics, any two elements of a vector space can be added to one another to yield another member of that space. This is one of the requisites of what it means for something to be a "vector" in mathematics. Specifically, a ...


5

Fun question. The muon density inside a white dwarf is negligible, because Fermi suppression does not really apply. Fermi suppression is the technical name of the effect you were describing: the decrease in the rate of a process due to the fact that there are no free states to accommodate one of the decaying particles (an electron, in this case). The ...


5

I think that it is only necessary to use the cyclic identity. Contracting both sides with the Levi-Civita, we should have $$0 = (R_{abcd} + R_{adbc} + R_{acdb}) \varepsilon^{abcd} \tag{1}.$$ Let $S = R_{abcd}\varepsilon^{abcd}$. Then $R_{adbc}\varepsilon^{abcd} = -R_{adbc}\varepsilon^{acbd} = R_{adbc}\varepsilon^{adbc} = S$ where the last step is renaming ...


4

Using cylindrical coordinates with the origin at the center and the $\phi = 0$ direction 'down' (the OP says the image should be rotated CCW 90 degrees), the electric field appears be have only a radial component with a sign change for $\phi = \frac{-\pi}{2}$ and $\phi = \frac{\pi}{2}$ $$\vec E = E(\rho,\phi)\hat\rho $$ $$E(\rho,\phi) = ...


3

I'm pretty sure I know the answer to this question, even though this question provides very little context for what the various tensors are, it gives a couple expressions without including the important surrounding equation for context, and it includes an equation with a typo. First of all, as it stands, the first equation has unbalanced indices. I assume ...


3

The time-independent Schrödinger equation for the hydrogen atom is $$-\frac{\hbar^2}{2m}\vec \nabla^2\psi-\frac{e^2}{4\pi \epsilon_0r}\psi=E\psi $$ If your aim is just to verify that the $1s$-wave function $$\psi_{100}=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}\hspace{2cm} a\equiv \frac{4\pi\epsilon_0\hbar^2}{me^2} $$ is indeed an eigenfunction, then your task ...


3

Can we add any two vectors? My book says it is true For example, adding acceleration to velocity (seems impossible). Quite simply, your book meant two vectors of the same type. It's just that simple. As you thought, you can not add two vectors that are "different things"! (If you're just getting started with vectors. Note that indeed vectors have a ...


2

I just figured out the answer on my own so I'll post it here for the sake of the site. $$\Delta x = v_0 t \cos \theta $$ $$\Delta y = v_0 t \sin \theta - \frac{1}{2}g t^2$$ Where $\theta, \Delta x, \Delta y$ are all known. Using the first equation, we get $v_0t = \frac{\Delta x}{\cos \theta}$ Now substituting into the second, we find $\Delta y = \Delta ...


2

Of course you can charge a capacitor with AC. The problem is that you keep changing how it is charged. While you apply a positive voltage to one plate, it will get a positive charge; half a cycle later, it will attempt to get a negative charge; and so it continues. The capacitor is always a little bit behind - as your AC voltage is changing, the capacitor ...


2

"I found out that F applied to mC must equal to the tension exerted on mA" is wrong F=(m_a+m_b+m_c)*a T=m_a*a T_x=m_b * a T_y=m_b *g T^2=T_x^2+T_y^2 Replace and get F=[(m_a+m_b+m_c)*m_b*g]/sqrt(m_a^2-m_b^2)


2

The differential equation is not complicated and has an analytical solution. Using $F=ma$ you get $m\frac{dv}{dt}=-a\cdot e^{b\cdot v}$, move both dt and the exponential to the opposite side and you get the solution for $v$ after integration of a simple exponential. Repeat using $v(t)=dx/dt$ to get the distance


2

Yes is correct. The rotation of the armature commutes the connection to the circuit so current in the circuit flows in the same direction regardless of the direction in the armature. If the armature is fixed, it will only couple to the circuit in a unique way, and the circuit will experience the current in the same sense as the armature, which is the way you ...


2

How fast would a sphere need to rotate for a dust speck at its equator to achieve balance between gravitational attraction and centrifugal force? If you do the math (equating $G M m / R^2$ to $m \omega^2 R$ and using $M = \frac{4\pi}{3} \rho R^3$ as well as $\omega = 2\pi f$), it follows that the size of the sphere is entirely irrelevant and that only the ...


1

This question when applied to the Earth is purely academic. The easiest solution is the one posted by Johannes: One revolution per one hour and 24 minutes. Why go beyond that? The question is academic. It's not academic when applied to asteroids. There's an interesting effect, the Yarkovsky–O'Keefe–Radzievskii–Paddack effect, aka the YORP effect, that can ...


1

You're just thinking 'upside down'. The 3 ohm resistor is just 1/4 the resistance of the 12 ohm resistor but, power is inversely proportional to the resistance. Thus, if you decrease the resistance by a factor of X, you increase the power by a factor of X. In this case, the resistance is decreased by a factor of 4 so the power is increased by a factor of ...


1

When the magnet is rotating and the coils are stationary, you have an AC generator unless you play tricks with brushes. This is because the net flux change in the coil after a complete revolution is necessarily zero - and thus the net voltage must be zero too. The sinusoidal waveform in your second sketch is therefore what you expect to see in a brushless ...


1

The angle between $a_1$ and $a_2$ should be $2 \pi/3$, not $\pi/3$. Your equation is then $$\left(\frac23 a\right)^2 + \left(\frac13 a\right)^2 + \left(\frac12 c\right)^2 + 2\left(\frac23 \frac13 a^2\cdot \cos(\frac{2\pi}{3})\right)=a^2.$$ This gives $\frac49 a^2 + \frac19 a^2 + \frac14 c^2 - \frac29 a^2 = a^2$ which implies $c^2/a^2 = 8/3.$


1

Think of $\lvert \psi \rangle$ as being written $a \lvert n \rangle + b \lvert n{+}1 \rangle$ -- it is just a linear combination of $\lvert n \rangle$ and $\lvert n{+}1 \rangle$ with (possibly complex) coefficients $a$ and $b$. Converting from a ket to a bra (i.e., finding the dual) distributes over addition and scalar multiplication, and it ...


1

This is my interpretation: If the pendulum has a period $\tau$ (2 seconds in this case) then the number of oscillations per day,$N$, is the number of seconds in a day divided by $\tau$: $$ N = \frac{86400}{\tau} $$ If the pendulum loses $T$ seconds per day (20 seconds in this case) then the number of seconds lost per oscillation, $\Delta\tau$, is: $$ ...


1

My interpretation of the question; specifically the "lost seconds per day" A pendulum is used to regulate a clock; in its original location, the clock keeps perfect time, as judged against some standard: WWV time signal, chronometer, etc... The clock is packed up, transported to a new location, set up and set running. The clock is set to the correct ...


1

If the drag force is being modeled as a linear function of velocity $(\vec{F}_D=-b\vec{v})$, then the problem is straightforward. The vertical force balance for a falling droplet is $$\Sigma F_y=mg-bv=m\dot{v},$$ which gives the following differential equation for the velocity: $$\boxed{\dot{v}+\frac{b}{m}v=g}.$$ In the limiting case of the maximum ...


1

what is the energy of a photon? how much has that energy changed after interacting with the electron? what is the formula for the kinetic ennergy of an electron? if the electron was not moving, and all of the change of the photon energy changed th eelectron's kinetic energy, then what is the speed of the electron? divide the answer from 4. by c.


1

I have came up with this: Charges are the sources of the electric field. So, whatever the point that field lines are "created" or "destroyed", must be a charge. Then, if there are a charge, then must be on the center. Calculating the electric flux: $$ \phi = \iint_S\ \mathbf E\cdot d\mathbf s = \frac{Q}{\epsilon_0} $$ Let's pick a sphere as gaussian ...


1

Integration by parts or using the identity $\frac{d}{dt}\dot{\phi}^2=\dot{\phi}\ddot{\phi}+\dot{\phi}\ddot{\phi}$ would be the trick here: $\frac{d\phi}{dt}dt=d\phi$, so: $\int \ddot{\phi} d\phi=\int \ddot{\phi} \dot{\phi} dt=\dot{\phi}^2-\int \ddot{\phi}\dot{\phi}dt$. Applying the usual trick (adding the integral to both sides and dividing by two), this ...


1

If I understand your question correctly you are asking why you should use $t_2-t_1$ in the denominator instead of $t_2+t_1$. The reason is that both measurements are taken with respect to some initial time. This initial time is arbitrary; think about it as the time when you started your stopwatch. The total time elapsed between the events occurring at ...


1

I think you are pretty close to understanding it. The "A" set of equations represent the components in the primed coordinate system shown with axes $x'$, $y'$. Here $y'$ is what's usually considered "vertical" (normal to the tangent plane to Earth where you happen to be , if you will). You are exactly right, $mg$ is not considered in Equation 1a because ...


1

Well I imagine you know how to add resistors in parallel: $$R = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}}$$ Of course if you try to put a resistor with $R=0$ in there you run into $\frac{1}{0}$. You can probably take a limit and show that the equivalent resistance is $0$ if there's a zero resistor in parallel. This makes sense. If current is ...


1

No need to involve time in your calculations. If the acceleration is constant, then we know purely from kinematics that $$V_f^2=V_0^2+2a\Delta x,$$ and in this case the final velocity $V_f=0$, so the magnitude of the acceleration is simply $$a=\frac{V_0^2}{2\Delta x}=\frac{F_f}{m}.$$ From the definition of the coefficient of kinetic friction $\mu_k$, we know ...


1

One approach to gaining insight to the answer is to note that, for two parallel resistors, the equivalent resistance is $$R_{EQ} = \frac{R_1R_2}{R_1 + R_2} = \frac{R_2}{1 + \frac{R_2}{R_1}}$$ as long as either resistance is not equal to zero. Now, let $R_2$ go to almost zero while holding $R_1$ constant. When $R_2 \ll R_1$, the equivalent resistance is ...



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