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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

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 ...


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

...truck will experience larger force. But Newton's Third Law of motion says that 'to every action there is an equal and opposite reaction'. So the force experienced by the truck (M) should be same to that experienced by the car (m), but negative, isn't it? Third law states that momentum must be equal and opposite if both vehicles must come to a ...


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 ...


2

Hint: You're far enough from the ground that the formula $mgh$, which is an approximation, doesn't really apply anymore. What is the true formula for gravitational potential energy?


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

Hint: Divide the $n$-sided regular polygon into $2n$ right-angled triangles. Calculate the moment of inertia for a triangle with the appropriate dimensions and appropriate mass rotating about the appropriate point, which will require a much easier integration than would calculating the moment of inertia for the whole polygon at once. Then multiply the ...


2

You calculated the specific potential energy at that distance. You were asked to calculate the potential energy needed to reach that point. You did two things wrong in that calculation. You forgot to multiply by 930 kg and you forgot to use the given condition "that the potential energy of a 930 kg object on the Earth's surface is -58.7GJ". What you need ...


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

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

There's an error in that the type of pipe for each of the two fundamental frequencies as described in your comment don't match the problem description. The pipe with a fundamental frequency of 440Hz is open-closed, and the pipe with a fundamental frequency of 660Hz is open-open. You actually said "closed-closed", which isn't even an option, but even if ...


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

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 ...


1

For equation 1a we don't take component of $ mg $ because it's component in this direction is zero (your guess is right) $$ mg\cos (90) = 0$$ The same thing happens for normal in 1a and $ T $ in 2b, since direction "x" is perpendicular to direction of normal and direction "y" is perpendicular to direction of $ T $ their components in respective directions ...


1

Find the integration by applying the limits in the sixth equation. Multiply by c. Substitute $P=cV^{-n}$ twice.


1

The answer depends on what you're trying to do. If you're trying to find the voltage between A and B, then note that the circuit can be redrawn as From this redrawn circuit, it's clear that the resistors need to be treated as being in series in the process of calculating the voltage between A and B. But if you're trying to find the Thévenin equivalent ...


1

No, block 1 would move to the right but with smaller acceleration than block 2. You could say it would move to the left relative to the block 2. Yes it would, since block 1 still 'wants' to move to the left relative to the block 2 (the 'relative' part is very important) No it wouldn't, it would be smaller. Block 1 would now move with acceleration that is ...


1

Original answer 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 ...


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

Assume the contrary, the volume of liquid displaced is greater than, or less than, the volume of the object. Then there is either a volume containing neither water nor object, or there is a volume occupied by both water and object.


1

They both experience the same force because of the impact, due to the Newton's third law, like you say. I think the question is not clear enough. If you assume there is no friction between the trucks and the ground, then you can use momentum considerations. I know this shouldn't be an answer, but I'm new and I can't post a comment, yet.


1

There is one error in the derivation, if you want to have $v(t_i)=v_0$, you must have $$v(t) = v_0 + a(t-t_i)$$ You also have to use the fact that $v_f = v(t_f)$. Once you use all this, you should be able to divide out $t_f-t_i$ in the corrected version of your last line and get the result you seek.


1

Yes, the force $A$ exerts on $B$ has to be equal to the one $B$ exerts on $A$. And when $F$ vanishes the forces between $A$ and $B$ vanishes too. More specifically, if in the time $F$ is acting no movement occurs (i.e. $F$ does not overcome the total friction of $A$ and $B$) the only thing causing $A$ and $B$ to interact is $F$, so this action reaction ...



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