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12

Just based on the quadratic drag of air, yes, the fired bullet would take longer to hit the ground. Just consider the vertical force caused by the air friction: $F_y = - F_{\rm drag} \sin \theta = - C (v_x^2 + v_y^2) \frac{v_y}{\sqrt{v_x^2 + v_y^2}} = - C v_y \sqrt{v_x^2 + v_y^2}$ Where $\theta$ is the angle above the horizon for the bullet's velocity, ...


8

You might get an order of magnitude estimate as follows. We make the rough assumption that everything ends up in its vessel as a monoatomic ideal gas - actually it will be a plasma, with a thermal energy per mole of $\frac{3}{2}\,R\,T_{final}$, where $T_{final}$ is the thermodynamic temperature of the plasma. Neglecting heats of vaporisation (we assume ...


5

If the car starts out going in a straight line, it will drop a little bit in the time it takes to cross the gap. If the drop is larger than the height of the chassis above the ground, the car will crash into the opposite wall. When the drop is less than that (small gap, high speed) and the wheels are able to absorb the shock, it is possible that the car will ...


3

The hand-wavy way to do it is to consider a wave solution like the one below, and apply Faraday's law to loop 1, and Ampere's law to loop 2: If you make the loops narrow enough, i.e., their widths are $dx$, then $$\oint_1\!\vec{E}\cdot \vec{ds} = -\frac{d\Phi_B}{dt} \to \frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{dt}$$ $$\oint_2\!\vec{B}\cdot ...


3

You can use the parallel axis theorem to work out the moment of inertia of a rod of length $l$ with it's centre of mass displaced from the axis of rotation by $\frac{l}{2}$ then multiply this value by four to get the moment of inertia of the whole square. The parallel axis theorem is: \begin{equation} I = I_{cm} + md^2 \end{equation} Where $I$ is the ...


3

When you push off, as you go forwards, the boat moves backwards. Momentum is conserved, so that your mass$\times$velocity equals the boat's mass$\times$ the boat's velocity. If you do the sums you will see that if you and the boat have comparable masses then the boat moves backwards quite quickly. Because you exert the force for your leap over a small, but ...


3

The answer depends on many factors, but here are the basic bits of physics that play: The power and wavelength of the laser The reflectivity of the surface (function of wavelength of the laser) The size of the focal spot The thickness of the sheet The thermal conductivity of the sheet The reflectivity of the copper is a particularly important one. If you ...


2

First, the obvious explanation for the sign is that if $J$ has a minus sign in (1), then there should be a minus sign in (4). For some reason your $G$ turned into $\phi_i$. Assuming that they are the same thing, then I'm not sure I understand your problem. We didn't use the homogeneous KG equation to get the delta function; we used the inhomogeneous one, ...


2

*elastic collision occur between atomic particles? inelastic collision occur between ordinary objects? perfectly inelastic collision occur during shooting? super elastic collision occur during explosion?* as John Rehnnie has explained, an elementary particle as the term implies, is not made of other particles, has no lattices ...


2

All your thinking is very good and correct. BUT you are missing a point in the question :) Yes, the door's weight (which pulls from the center og gravity - no need to think about each particle of the door) creates a torque. The hinges then gives a counter torque. True. BUT let's read the question: A door is hinged at one end and is free to rotate about ...


2

The error in the OP's question comes from the second Kronecker delta, which is not correct (furthermore, the sum over $k$ is not dealt with...). After using the first Kronecker delta $\delta_{k\,k'}$, one has to use the identity $$\frac{1}{N}\sum_{k=0}^{N-1}\exp{[\frac{2\pi i k(j'-j)}{N}]}=\delta_{j'\,j},$$ which directly gives that $F^\dagger F =1$. One ...


2

The first one is correct, $L_{new}=2L$. In your second expression $L=\frac{v^2}{2a}$, when changing $a$ you must also find a new end speed $v$, $$L_{new}=\frac{v_{new}^2}{2a_{new}}=\frac{1}{2}\frac{v_{new}^2}{2a}\neq\frac{1}{2}\frac{v^2}{2a}=\frac{1}{2}L$$ A note from reading the comments to the question: I am assuming the objects fall during the same ...


2

Your mistake is that the two tensions are different, because of the presence of a pulley with non-zero $I$. What you have missed is: 1) connect the two tensions to the torque: $(T_1-T_2)R=I\alpha$ and 2) link the accelerations $R\alpha=a_1=a_2$ NOTE (from comments): If the pulley had a zero $I$ (moment of inertia), then the two tensions would be be ...


2

The tension in the rope should be different on the left than on the right - it is this difference that gives rise to the torque that accelerates the pulley. You seem to think that it should be the same: but if it was, then where would the torque to move the pulley come from? Annotate your diagram carefully: you did not show $T$ anywhere.


2

The center of mass of an object is the point where the first moment of mass is zero. Put differently, when you support the object at that point, it will be balanced. Assume that point is $x_0$, then $$\int_0^\ell (x-x_0) \lambda(x) dx = 0$$ Substitute $\lambda$ and some simple manipulation will give you an expression for $x_0$. Let us know how far you ...


2

In the context of a canonical transformation (CT) $$z^I~=~(q^i,p_i)~\longrightarrow ~(Q^j,P_j)~=~Z^J~=~f^J(z,t),$$ the matrix $$\textbf{M}^J{}_I~:=~\frac{\partial Z^J}{\partial z^I}$$ is the Jacobian matrix of the CT. Here the indices $$i,j~\in~\{1,\ldots, n\} \quad\text{and}\quad I,J~\in~\{1,\ldots, 2n\}.$$ If the CCR reads $$ ...


2

That's because you want $F_1 + \Delta F = F_2$ by additivity of vectors (for a more rigorous approach, see the formalization of affine spaces). Hence, $\Delta F = F_2 - F_1$ PS: I couldn't comment because of my low reputation, so I made an answer for so little


2

I am not going to solve this completely for you, but note just a few things: The charge of an electron is negative In a simple harmonic oscillator, the force must be in a direction opposite to the displacement You had earlier set $x=A\sin\omega t$ so it seems to me that $\frac{x}{A\sin\omega t}=1$ I think that if you ponder the above you will see where ...


2

The differential and integral forms of Maxwell's equations are truly equivalent; they are essentially the same set of equations. One can convert between the two using two mathematical theorems: Divergence Theorem (Wikipeda - Divergence Theorem) Stokes' Theorem (Wikipedia - Stokes Theorem) The divergence theorem states that the flux over a closed surface ...


1

$\newcommand{\G}{\mathbf{G}} \newcommand{\H}{\mathbf{H}} \newcommand{\A}{\mathbf{A}} \newcommand{\B}{\mathbf{B}} \newcommand{\tH}{\tilde{\H}} \newcommand{\tG}{\tilde{\G}} \newcommand{\Hp}{\H^+} \newcommand{\Gp}{\G^+}$I will prove the answer two ways. The first way is the way you were "supposed" to do it, and the second way is another way of doing it. The ...


1

The other answers are OK, but if I'm correct they are missing information. Firstly, to be completely thorough, a general approach to force questions is to split the forces into components as shown here. If you do that and add the vertical force components and the horizontal force components, you will get a net force. This net force is the direction of ...


1

$F1 - F2$ is the same as turning $F2$ around 'head to tail' and you get $-\Delta F$ or a vector, which is the same magnitude, but the opposite direction to $\Delta F$. To get $\Delta F$ you need to reverse $F1$ as in the diagram I have modified from your diagram below


1

You can try easily what is the good answer by calculating the difference in energy between the two participating levels. $E_{upper} - E_{lower} = -Rhc [\frac {1}{(n+1)^2} - \frac {1} {n^2}]$ $= Rhc [\frac {2n + 1}{n^4 + 2n^3 +n^2}]$ So, its obvious that the smaller n, the bigger $E_{upper} - E_{lower}$.


1

Although I think you may get the solution already, let me write answer as I thought. First, as you wrote the energy $E$ can be written as \begin{equation} E=m(-\xi_\mu\xi^\mu)^{1/2}=mV. \end{equation} Because $E$ is conserved, $V$ is also conserved. Second, we are now in asymptotically flat spacetime. This means that at infinity you get $V=1$. Therefore ...


1

You can calculate the work done by gravitational force as the product of its weight and y-displacement. If I have got your question right, the body is freely falling after the force tips it off the table. So the work done by your force will not be as you have written. It would've been correct if the force had been acting on the body throughout its ...


1

The idea is that since the steel beam has conduction electrons that are free to move, the movement of the charges in a magnetic field causes a magnetic force to act. The magnetic force causes the electrons to accumulate at one part of the curved surface of the rod, thereby creating a potential difference. The charges keep accumulating till the potential ...


1

Consider the following results: From the definition of scalar product of four vectors, $$ \tag{1}(p_1 p_2)^2 \equiv (p_{1\mu}p_2^\mu )^2 = (E_1E_2 - \textbf{p}_1 \cdot \textbf{p}_2 )^2.$$ The usual dispersion relations: $$ \tag{2} E_i = \sqrt{ | \textbf{p}_i |^2 + m_i^2}.$$ The velocity $\textbf{v}_i$ in terms of momentum and energy: $$ \tag{3} ...


1

In the Schwarzschild geometry, the Schwarzschild radius breaks naive dilation symmetry. In the simple case of a radial dilation $r \to \lambda r$, the geometry is only preserved by $R_S \to \lambda R_S$. So, it naively seems like it would be difficult to find a working dilation, even just a radial dilation. I went to some effort (as an exercise for myself) ...


1

Since the surface is frictionless there is only vertical force. The torque is given by the normal force of the surface multiplied by the horizontal distance to the center of mass (c.o.m.). Now the normal force depends on the vertical acceleration of the c.o.m. - you know that the acceleration of the c.o.m. is a result of all the forces acting on the object, ...


1

Unlike in QFT where you can derive spin from more basic principles, in ordinary non-relativistic QM spin is essentially defined into existence as a group of operators $S^i = (\hbar/2) \sigma^i$ that satisfy the algebra $$[\sigma^i, \sigma^j] = i \epsilon^{i j}_{\,\,k} \sigma^k.$$ The dimensions in the Hilbert space on which the Pauli operators act are ...



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