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45

I am sorry to say, but your colleague is right. Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity, $$ m g = k v^2 $$ And thus $$ v=\sqrt{\frac{mg}{k}}$$ So, the terminal velocity of a ball 10 times as ...


19

Ball 1 will drop faster in air, but both balls will drop at the same speed in vacuum. In vacuum, there is only the gravitational force on each ball. That force is proportional to mass. The accelleration of a object due to a force is inversely proportional to its mass, so the mass cancels out. Each ball will accellerate the same, which is the ...


11

Other answers & comments cover the difference in acceleration due to friction, which will be the largest effect, but don't forget that if you are in an atmosphere there will also be buoyancy to consider. The buoyancy provides an additional upward force on the balls that is equal to the weight of the displaced air. As it is the same force on each ball, ...


10

tl;dr: Velocity required: 1680 m/s Time to hit you: 6500 seconds Part 1: Velocity required (Using Google search values) Radius of moon = 1737.4 kilometers Mass of moon = 7.34767309E22 kilograms Assuming perfectly circular motion of the bullet, and no air resistance, and ignoring gravitational effects of other planets / objects in space, and using simple ...


6

The period of an elliptical orbit is given by: $$ T = 2\pi\sqrt{\frac{a^3}{GM}} $$ where $a$ is the semi-major axis. For a circular orbit of radius $r$ we have $a = r$. The two orbits you show do not have the same semi-major axis, so they do not have the same period. However if the elliptical orbit had $a^3 = 4r^3$ then the period of the elliptical orbit ...


5

An interesting way to answer Part 2: Using the angular velocity version of the centripetal force equation:$$F_c=m\omega^2r=\frac{GMm}{r^2}$$If we assume that $r$ is both the orbital radius and the radius of the sphere being orbited and that the density of that sphere is $\rho$, then:$$M=\frac43\pi \rho r^3$$ then the equation ...


5

$$ 110 \text{ hp} = 82 \text{ kW} $$ This is 1000x what the laptop draws. You won't notice. To put that change in perspective, you would see a similar increase in the power (85 W) used by the car if your speed changed from 65.00 mph to 65.02 mph, since $P \propto v^3$ (at high speed, the power goes as the velocity cubed) as per this answer, so $ ...


4

While people normally quote Newton's Second law as $\vec F = m \vec a$, it is better written as $$ \vec F = \frac{d\vec p}{dt} $$ Force is a rate of change in momentum. This means that the average force applied when an object undergoes some discrete change in its momentum is $$ F_{\text{avg}} = \frac{\Delta p }{\Delta t} $$ The change in your momentum ...


4

Your friend is correct. The acceleration of the projectile is determined by the thrust its rocket motor can produce. If the acceleration of the projectile is $2g$ then the thrust of its motor would be $2mg$. But launching the projectile from your space vehicle can't increase it's thrust. You can increase its initial velocity, but once the projectile has ...


4

The time evolution operator of a quantum system is (in units with $\hbar = 1$) $$ U(t_0,t) = \mathrm{e}^{\mathrm{i}H (t - t_0)}$$ and the "stationary states" are the eigenstates of this operator, i.e. eigenstates of the Hamiltonian. If you are given a collection of stationary states (not a basis of the space, mind you) $\{\lvert \psi_E \rangle\}$ with $H ...


4

The integral $$I(k) = \int_{-\infty}^\infty \frac{s e^{isr}}{(s-k)(s+k)} ds \tag{1}$$ where $k$ is real and the integration is for real $s$, is not really well-defined. This is precisely because the integrand has singularities on the integration domain. However consider if $k$ is a complex number $k = k_r + ik_i$ with $k_i >0$. Then the integrand is ...


4

Hint: Recall that $$\tag{A} q\cdot (q+2p) ~=~(p^{\prime}-p)\cdot (p^{\prime}+p) ~=~p^{\prime 2}-p^2~=~m^2-m^2~=~0 .$$ So $$\tag{B} \Delta -(1-z)^2m^2~=~y(y-1)q^2 -2yz q\cdot p~\stackrel{(A)}{=}~y(y-1+z)q^2~=~-xyq^2. $$


4

The wind is certainly doing work, because it applies a force and the point where the force is applied is displaced. However it isn't doing any work on the boat, it's doing the work on the water. The key point is that the net force on the boat is zero. We know the net force on the boat is zero because the boat is moving at constant velocity - if the net ...


4

The distance travelled in a time $t$ is: $$ s = ut + \tfrac{1}{2}at^2 $$ So the distance travelled between $t$ and $t - \Delta t$ is: $$\begin{align} \Delta s &= s(t) - s(t - \Delta t) \\ &= ut + \tfrac{1}{2}at^2 - u(t - \Delta t) - \tfrac{1}{2}a(t - \Delta t)^2 \\ &= u\Delta t + \tfrac{1}{2}a(2t\Delta t - \Delta t^2) ...


4

The faster you go, the less velocity you theoretically can gain from a gravity assist. The reason for this is that the faster you go the harder it is to bend the orbit. To proof this we have to use the patched conics approximation, which means that while within a sphere Kepler orbits can be used. The sphere can be simplified to be infinitely big, since the ...


4

By analogy (between $\mathbf{E}$ and $\mathbf{B}$ as they are pretty much equivalent) then the divergeance of $\mathbf{B}$ field wouldn't be 0 anymore, instead: $$\nabla \cdot \mathbf{B}= \frac{\rho_{\rm magnetic}}{\mu_0} $$ With $\rho_{\rm magnetic}$ the magnetic charge density, and $\mu_0$ the permeability in vacuum, to interpret it, the divergence of the ...


4

Consider the following diagram: This shows a mass $m$ moving past a point $P$ in a straight line. Note that the mass isn't connected to $P$ in any way - it's just moving past in a straight line. The angular momentum of $m$ about $P$ is given by: $$ \vec{L} = \vec{r} \times m\vec{v} $$ So the direction of $\vec{L}$ is normal to the screen and the ...


3

I think it might be this $$mgh=\frac{1}{2}mv_{cm}^2+\frac{1}{2}I_{cm}\omega^2=\frac{1}{2}I\omega^2.$$ I mean, it seems that you added the kinetic energy due to center of mass velocity to the rotational kinetic energy respect to the rotational point. It should have been respect to the center of mass.


3

I did not check thoroughly your free body diagrams,but they look correct. One comment though, I do not think that using polar separation of the accelerations is particularly useful for this problem since the obvious origin for such system is accelerating, as you well indicate. I think that your missing equations are just your geometrical constrains. You ...


3

Looking at even indices, from $a_{2j_0+2} \approx \frac{1}{j} a_{2j_0}$, we get : $$a_{2j_0+2k} \approx \frac{(j_0-1)!}{(j_0+k-1)!}a_{2j_0} = \frac{1}{(j_0+k)!} C \tag{1}$$, with $C=(j_0+k)\,(j_0-1)! \,\,a_{2j_0}$ So, taking $n= 2(j_0+k)$, we have finally $a_{n} \approx \frac{1}{(n/2!)} \, C$ Finally, $h_{even}(\xi) \approx \sum\limits_{n= j_0 + ...


3

I think Jerry's point is that if you have a metric that looks like: $$ ds^2 = a(dx^0)^2 + f(dx^1, dx^2, dx^3) $$ where $a$ is a constant and $f$ is any function, then an observer moving in the $x^0$ direction follows a geodesic of the type you describe. An obvious example is the FLRW metric where a static observer follows the geodesic $t = \tau$, $x = y = ...


3

One can get an order of magnitude estimate of the maximum speed attainable by gravitational slingshots without doing any real calculation. The 'rough physics' reasoning goes as follows: The gravitational field of the planets used for slingshots needs to be strong enough to "grab" the speeding spaceship. As a planet cannot "grab" a spaceships moving faster ...


3

It is a perfectly well-defined expression because the tensor product is a linear space. The vectors $|v\rangle\otimes |w\rangle$ form a basis of the whole tensor-product vector space, so any vector (including Bell's state) in this space may be written as linear combinations of such basis vectors. $$ |\psi \rangle = \sum_{ij} c_{ij} |v_j\rangle\otimes ...


3

For any curve in two dimensions described by a position $(x(t), y(t))$ as a function of time, the speed of such a curve at a time $t$ is \begin{align} |\mathbf v(t)| = \sqrt{\dot x(t)^2 + \dot y(t)^2} \end{align} where overdots here mean derivatives with respect to time $t$. To find the total distance traveled along the curve from a time $t_a$ to a time ...


3

This is really just a footnote to alemi's answer. The electricity for your car is supplied by the alternator, and the torque required to turn the alternator depends on the current it's supplying. As you draw more current more torque is required to rotate the alternator and the car has to use more power to do it. So yes, plugging in your laptop will increase ...


2

I'll assume you want to know the equilibrium points. The Lagrangian tells you everything you need to know about the system. Because variation of generalized momentum is: $$ \frac{dp_k}{dt} = Q_k + \frac{\partial T}{\partial q_k} = -\frac{\partial V}{\partial q_k}+ \frac{\partial T}{\partial q_k} = \frac{\partial L}{\partial q_k} $$ Then: $$ ...


2

Perhaps the best way to understand this is to start simply: Consider a function $f(x)$. Now, let's try to take the Fourier transform of its derivative $f'(x)$. Just use the definition of the Fourier transform: $$\mathscr{F}(f'(x))(k)=\frac{1}{\sqrt{2\pi}}\int dx\ e^{-ikx}f'(x) $$ and now use integration by parts (assuming $f(x) \to 0$ as $|x|\to \infty$, a ...


2

If we say we have two objects, with in general charge $Q_1$ and $Q_2$, which experience a certain electrostatic force $F$ between them, then we know that if we double the charge on one, we will double the force - that's just how electrostatics works. Force is proportional to the charge on each of the two objects. With that insight, we can say that the ...


2

First of all, the first equation for $ds^2$ is only valid if $f$ is nothing else than the azimuthal angle $\phi$. Second, if you are evaluating $X_i X^i$, the squared distance from the origin without any infinitesimals, then it is exactly equal to $-t^2+r^2$ and nothing else. The polar coordinate $r$ is chosen as $\sqrt{x^2+y^2}$ so its square already ...



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