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8

. After the collision, according to the conservation of energy, the speed of both balls should be 35 m/s. ... Now, obvisouly, 35 m/s in R2 is 85 m/s in R1, not 79m/s. Where does the discrepancy come from? The collision is an artificial example: it is considered inelastic, but there is no loss of energy. If it followed the ordinary rules, in ...


7

The stroke of genius The laws of Nature are wonderful: extremely simple - make a tremendously complex world, a few symbols can be extremely powerful and beautiful all roads lead to Rome, you may go from London to Rome following many different routes, only a fool would get there via Moscow. But an eagle can fly over the Alps in a straight-line. I'll try to ...


6

After amplification (more sound wave!) by outer hair cells in your ear, the sound wave ends up moving the flimsy bits of inner hair cells, a sensory input of your inner ear, expending some energy in the process. The rest of the energy in the sound waves also gets converted into something else, namely heat as it interacts with various parts of your ear.


5

(a) Write down the potential energy of the rope as the function $y(x)$. You're almost right, up to a minus sign in the limits of the integral: $$V=\dfrac{mg}{l}\int_{-x_0}^{d-x_0}y(x)\sqrt{1+y'^2}\ dx$$ (b) Since the problem is static, interpret the potential energy as the Lagrangian and find the Lagrangian density. The Lagrangian is usually given ...


4

The sound waves die down. Obviously, the sound waves, which cause a hearing sensation in our ears, cause physical movement of the eardrum and the like. but I honestly can't come up with anything that makes sense, because obviously, energy cannot be created nor destroyed. Don't worry, energy is NOT being destroyed here, it is just transferring from ...


3

You were pretty close already. There is a handy table on Wikipedia for a variety of coordinate systems. But for the polar system: $$ \vec{\nabla} \cdot \vec{U} = \frac{\partial U_r}{\partial r} + \frac{1}{r} \frac{\partial U_\theta}{\partial \theta} $$ and you can look up the curl in the same table. These can be derived from the Cartesian definitions by ...


3

The total force acting on a raindrop equals $g$ minus air resistance which increases with velocity. In other words, as the raindrop speeds up, air resistance increases which decreases the acceleration (until eventually the acceleration equals zero and terminal velocity has been reached).


3

The ear picks up only a tiny fraction of the energy in a sound wave, but does so very effectively because of the acoustic arching mechanisms in the inner ear. The sound energy (pressure) is converted into motion of cells, and thence into heat. The amount of heat is miniscule: the sound level in a "quiet" room might be around 40 dB which corresponds to an ...


3

Since he is falling at terminal, hence constant, velocity, he is experiencing NO net (total) force. There are (at least) 2 forces acting on him though, which are his weight (900N) and the air resistance. Because he has constant velocity, i.e. he is not accelerating, these forces must balance, i.e. the air resistance is 900N upwards whilst his weight is 900N ...


3

Suppose I drop a ball from a height $H$ above a flat ground, how can I determine how far it will have traveled by the time the ball hits the ground for the 3rd time $L=x$?. I know how to calculate the speed $V$ when it first hits the ground, using constant acceleration formulas, I can also work out how fast the ball is travelling ($v$), after ...


3

On an intuitive level, the initial speed $v_0$ can be considered to have two effects: one on the horizontal velocity, and one on the vertical; the former affects the range in a direct sense, and the latter increases the time the projectile is in the air. The combination of both of these gives an overall $v_0^2$ contribution. If you were to increase the ...


2

You forgot to multiply $T \sin{\theta}$ by the distance from the wall to the end of the bar in the torque balance. When you do that, you get an extra factor of 4 in the first term for the expression for x, $x = \frac{8\sin{\theta}}{\mu_s \cos{\theta} + \sin{\theta}} - 2$, which is positive. (PS: I didn't check your math, I just added the factor of 4, so I ...


2

It is an edge dislocation. Compare: to: TEM tracks dislocations in graphene Notice that the yellow loops have 5 and 7 edges, respectively, compared to the usual six.


2

I'll do all calculations assuming the lagrangian $\mathcal{L}$ acts on a 1-dimensional manifold $M$. I believe you'll find the generalization absolutely trivial, and this will spare me of writing tons of sums. Let \begin{equation} \mathcal{L}: \mathbb{R} \times T M \rightarrow \mathbb{R} \end{equation} be a lagrangian over $T M$, with time in ...


2

the speed of both should be according to the conservation of energy If both balls have the same speed after the collision, the collision is inelastic, i.e., kinetic energy is not conserved. If the balls are identical, then conservation of momentum requires that $$\mathbf v'_1 + \mathbf v'_2 = \mathbf v_1 + \mathbf v_2 = 150 \mathrm{\frac{m}{s}}$$ If ...


2

Once the mass is released, the center of mass will move at a constant velocity. Superposed on that is the relative motion of the two masses - first towards each other, then away. They will be in exact antiphase so the center of mass has constant velocity. Your mistake was to set x up as a cosine function - that implies that it is at an extreme of position ...


2

If one particle is fixed, some force is keeping it fixed, and in the presence of an external force, conservation of momentum doesn't apply. Your second equation is then $v_1 = 0$ (assuming particle #1 is the one that is fixed), not $m_1 v_1 + m_2 v_2 = 0$.


2

First, let's review the basic ideas of simple harmonic motion (I'm assuming an early university level). Starting with Newton's equation: $$F=ma$$ and using Hooke's law $$ma=-kx$$ then recognizing that acceleration is the second derivative of position x $$mx''= -kx$$ We know that simple harmonic motion is sinusoidal, so we substitute $x=\sin(\omega t)$ ...


1

In the first case,momentum is conserved because force is applied to each charge from WITHIN the system.So,the center of mass of the system is constant.In the second case,in order for just one charge to move,it has to be put in an EXTERNAL electric field.So,you can see that momentum within the system which consists only of one charge can not be conserved.If ...


1

$m_1 = m = 10$ kg $m_2 = 2m_1 = 20$ kg $F = 60$ N; $ma = F - \mu mg-T$ $2ma = T - 2\mu m g$ $\Rightarrow ma+\mu mg = F-T$ $ 2(m a+ \mu mg)=T$ $\Rightarrow 1/2=(F-T)/T \Rightarrow T = 2F/3$


1

Suppose I drop a ball from a height $H$ above a flat ground, how can I determine how far it will have traveled by the time the ball hits the ground for the 3rd time $L=x$?. It is very simple, supposing the $C_R=.56$ At each bounce will travel $l = H * C_R^2$, and if you substitute $H $ with $h_1,h_2...$ at each bounce you get the formula for the ...


1

Suppose you and a friend shoot pool balls dead-on, which then bounce off of each other. Right before the collision, you count up the kinetic energy, and you measure it to be $E_1$. Right after the collision, you count up the kinetic energy, and you measure it to be $E_2$. There are clearly three possible cases: $E_1 = E_2$: This is an "elastic" collision. ...


1

What happened with $V\left(\sqrt{x^2+y^2+z^2}\right)$? You mean, why does V(r) disappear from the $\frac{\partial }{\partial \dot q_j}$ term, right? It's because V(r) is a function only of $q_j$ not $\dot q_j$. Those variables are treated as independent and so $\frac{\partial V}{\partial \dot q_j}=0$. and why $\partial\dot q_{j} = \partial\dot ...


1

To make the correct answer clearer, allow me to introduce the canonical momentum $\vec{p}$, given by: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}$$ This way we can rewrite the Hamiltonian as: $$H=\vec{p}\cdot\vec{\dot{x}}-L$$ Let's start by computing $\vec{p}$: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}=m\vec{\dot{x}}+\dfrac{e}{c}\vec{A}(\vec{x},t)$$ ...


1

$F=\frac{Gm_1m_2}{r^2}$ is wrong because you are essentially summing up the magnitude of the gravitational forces of each point on the ring without considering their direction. This is wrong since force is a vector. To find the force, take 2 small elements, diametrically opposite on the ring each of length $dl$ and mass $dm$. Draw the direction of the force ...


1

Calculating the gravitational force on the axis of a ring is equivalent to calculating the gravitational force of a pair of opposing small portions of the ring in which the full mass $M$ of the ring is thought to be concentrated. The result is an axial force equal to $$F = G M m\frac{cos(\theta)}{S^2}$$ Where $\theta$ is the half-angle between the two ...


1

Instead of integrating the force due to each mass element, which requires you to compute the component in the x-direction, you can calculate the gravitational potential, which is a scalar quantity. The force is then minus the gradient of the potential. The gravitational potential energy is: $$V(x) = -\frac{m_1 m_2 G}{\sqrt{x^2 + R^2}}$$ The force in the ...


1

You asked a similar question on worldbuilding for a story. Hit youtube for Operation Crossroads, Baker test. The US Navy detonated a 21kt device 90 feet underwater. Produced a nice fountain but no overly-destructive wave action after a few kilometers. A few years later, Castle Bravo (15Mt) also failed to produce any significant damage* outside the ...


1

There are two things going on in the last step. Let's take care of the simpler one first. The third term reduces thanks to a trigonometric identity: $$ -\hbar^2 \left ( 1 + \frac{\cos^2 \theta}{\sin^2 \theta} \right ) \frac{\partial^2}{\partial \varphi^2} = -\hbar^2 \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta} \frac{\partial^2}{\partial \varphi^2} = ...


1

Well,the calculation is correct if we suppose that Ra and Rb are correct. BUT,Ra and Rb are incorrect.The sum of Ra and Rb is equal to 110kN(you put Ra and Rb in N-correct that) but the sum of all the forces except the 0.5N/m is equal to 110KN.So,you must have forgot to put the 0.5KN/m in your shear force calculator.



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