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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

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

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

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

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

$\newcommand{\G}{\mathbf{G}} \newcommand{\H}{\mathbf{H}} \newcommand{\A}{\mathbf{A}} \newcommand{\B}{\mathbf{B}} \newcommand{\tH}{\tilde{\H}} \newcommand{\tG}{\tilde{\G}} \newcommand{\Hp}{\H^+} \newcommand{\Gp}{\G^+}$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 ...


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

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 ...


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

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

*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 ...


1

If pulleys are not weightless and maybe rope is not weightless. Weight $Q_1=m_1 g$ accelerates whole setup. Tension $T_1$ is responsible for acclereating two pulleys, rope and mass $m_2$, $T_2$ accelerates one pulley less to the same acceleration, so that it is lower. By the same reason $T_3$ is lower than $T_2$.


1

I think we are supposed to assume that the buoyancy force of the balloon is equal to the weight force of the balloon, ladder, and climber. If this is the case, the system is in equilibrium with its environment, with no net forces to the environment. You could look at it as a center of mass problem. We assume the ladder has negligible mass. When the climber ...


1

Well it all depends on what assumptions you are able and willing to make. For a start you can work out how much of the laser light is reflected. The reflected power will be something like (for normal incidence) $$\frac{P_{r}}{P_i} = \frac{(\eta_m - \eta_{vac})^{2}}{(\eta_{vac} + \eta_m)^2},$$ where $\eta_{vac}=377$ Ohms and $\eta_m$ is the impedance of the ...


1

From the equation given for terminal velocity v=$$\sqrt{\frac{2mg}{\rho AC}}$$ the density of the planet in question will be very low , so any object dropped will take a very long time to acquire a terminal velocity. There are so little particles present in the atmosphere that the balancing of the weight by the drag and buoyancy will take a very long ...


1

The air density of your planet is on the order of $10^{-18}$ of air at the surface of the earth. Gravity Force would be 10 times larger, even accounting for a slightly larger radius. The largest possible parachute that could fit around such a planet would be in the order of $10^{12}$ times bigger than a normal parachute. But it would need to have a mass in ...


1

Draw the position of the frame a very small time later. From this you can determine the relative rotation (due to torque - the difference between the forces multiplied by their perpendicular distance to the center of gravity) and linear acceleration (due to net horizontal and vertical forces). That should give you all the equations you need and you can then ...


1

One thing you need to consider is the kinematics. That is what is the position, velocity and acceleration of the center of mass G as the frame rotates. I set the coordinate system origin where the two slots intersect to get $$\begin{aligned} \vec{r}_A & = (x_A,0,0) & \vec{r}_B & = (0,y_B,0) \end{aligned} $$ The I recognized that the two ...


1

Thanks laying out your work so neatly in the question. I think the solution is the following $$\Delta KE= \int_{r_a}^{r_b}{ KQq \over r^2} dr$$ where $r_a$ is the initial position and $r_b$ is the final position (and I have added $q$ as the charge of the point charge). so, for example, if the point charge goes from $r$ to $2r$ we have two positive ...


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, ...



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