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7

Visualization The difference in height $h$ is always the same (here 10 m)! Remark This is of course only true if $g$ is constant, e. g. $h$ does not change "much". See also Wikipedia.


5

Potential energy is given only as a difference of energies at different heights. So, if you want to know just how much does the person gain energy (or rather loose by friction in their muscles and joints) by walking down the hill, you might just use their height of their heels on the top of the hill and under the hill. But remember, you always have to use ...


4

Theoretically, their center of mass is what you're looking for. It's somewhere near the stomach. High jumpers bend their body when they are jumping so that their center of mass can travel just above the bar which allows them to use the least energy to jump the highest: https://en.wikipedia.org/wiki/High_jump When a body is rotating or something unusual is ...


2

Consider the angle between the two vectors as $\theta$ and the following rules $$\begin{align} \vec{a}\cdot\vec{b} &= \|\vec{a}\| \|\vec{b}\| \cos\theta & \|\vec{a}\times\vec{b}\| & = \|\vec{a}\| \|\vec{b}\| \sin\theta \end{align} $$ Now to construct the parallel vector use the direction of $\vec{b}$ and the adjacent side of the triangle ...


2

Lets to this step by step and take care of the signs! Let $q_1=3$µC, $q_2=5$µC and $q_3=-8$µC. The formula for the force, acting on particle one due two the presence of particle two, is given by $$\vec{F}_{12}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{{| \vec{r}_{21}|}^2} {\hat{r}}_{21},$$ where $\hat{r}_{21}$ is the unit vector pointing from charge two to ...


2

At the risk of sounding like a broken record... it is a good idea to draw a diagram for all but the very simplest problems (and even then): You can immediately see that the normal force is made up of two components: $F_c \sin\alpha$ and $F_g\cos\alpha$. The friction results from the combination of both of these. In your approach, you ignore the normal ...


2

The binomial expansion says that $(1+x)^n=1+{n \choose 1}x^1+{n \choose 2}x^2 + ...$. This should be familiar to you for positive, integer n just by expanding out the parenthesis. For NEGATIVE n, it still holds, provided you interpret ${n \choose k}$ correctly for negative numbers; for our purposes, we just need to know ${n\choose 1}=n$ always. For very ...


1

Taking the Hermitian conjugate reverses the order of the $\psi$'s. You have $$ L_M^\dagger = \left( \bar{\psi}\psi\right)^\dagger = \left( \psi^\dagger \gamma^0\psi\right)^\dagger = \psi^\dagger{\gamma^0}^\dagger \psi = \psi^\dagger\gamma^0\psi = \bar{\psi}\psi = L_M \ , $$ where we use that $\gamma^0$ is Hermitian.


1

If this is a correct description of what happens, can we conclude that g does same work on P and on P'? Yes. This is correct. If g acts perpendicularly to the velocity, it performs work of magnitude zero. This is also correct. The reason the two statements above are not contradictory is that the work done by the gravity changes the direction of ...


1

yep, think of $ \xi $ as a unit vector and replace all instances of it with $ \epsilon \xi$ where $ \epsilon $ is some small number. Then you will see that those two terms are second order in $ \epsilon $.


1

As the value of $b$ increases the resistance between the outer and the inner shells will converge to $1/4 \pi \sigma a$. If we consider the outer shell to be at the "infinity", the resistance between the "infinity" and the inner shell will be $1/4 \pi\sigma a$. We can think of the situation in which there are two shells in the infinite sea of poorly ...


1

While not necessary to solve this problem, I want you to know that... Concept # 0: the angular velocity of circular motion is directly proportional to the linear velocity of motion, $$ v = \omega r $$ where $v$ is the linear velocity, $\omega$ is the angular frequency, and $r$ is the radius of circular motion. Concept # 1: whenever an object exhibits ...


1

If it is static friction, then the two blocks are stuck together and both have the same acceleration. In that case, the top block has a net force of 40 N (100 N pull - 60 N friction) and the bottom block has 60 N (just from the friction). Since 60N < 80N max friction, then the ansatz that this is static friction is consistent.


1

The problem is that you have not solved the question yet. What you have found is not the friction between the boxes. It is something else. As you actually state yourself, you have instead found the maximum [static] friction. This is just the maximum possible value and not at all necessarily equal to the actual friction. Static friction can be anything from ...


1

The problem is arising since you are trying to take the trace in a non-orthonormal basis including both $|e_1\rangle$ and $|e_2\rangle$. If $A$ is an operator, the trace of the matrix $$A_{mn}\equiv \langle e_m|A|e_n\rangle$$ is not invariant under any basis $|e_m\rangle$. If this is the orthonormal eigenbasis, the trace is the sum of the eigenvalues. Any ...


1

The axis of simultaneity, or in other words, the set of events which are simultaneous as measured in the rest frame of the ship, does indeed change suddenly when we turn back. This is because it depends on your reference frame. There isn't a single inertial frame that stays with the ship for the whole journey; you can either accept that the frame is ...


1

Perhaps it's more illuminating to look at the whole thing in a spacetime diagram. we have the earth frame with coordinates $(t,x)$, and its trajectory through spacetime is the blue line. The trajectory of the spaceship is the red one. Straight worldlines are inertial frames of reference, curved or non-straight worldlines are non-inertial frames of ...


1

I propose redefining this problem as follows (because I'm not sure it has a solution the way the OP has defined it). Let $y=f(x)$ be some symmetrical (around $y$) function like $x^2$. Let the point mass experience a friction force acc. to the usual simple model $F_f=\mu F_N$, with $F_N$ the Normal force acting on the point mass in the point $(x,y)$ ($N$ ...


1

The Laplace equation $\nabla^2 \psi = 0$ is a linear differential equation. Now note that if $\phi$ is real, then so is $\nabla^2 \phi$. Moreover, by the linearity of the equation, if $\phi$ is real, then $i\phi$ is pure imaginary, and so is $\nabla^2(i\phi) = i \nabla^2(\phi)$. Okay, back to your situation. Let's say the solution is $\phi_1 + i\phi_2$ for ...


1

First you half fill the pipette with hot water. So at that point it's filled half with hot water in the lower part, half filled with air at room temperature in the upper part. Then you block the top end of the pipette with the tip of your finger and turn the pipette upside down, so the hot water starts flowing down to the blocked side. However, the presence ...


1

for same voltage supply, the power consumed by two resistances in series connection is less in compare to power consumed by same resistances in parallel connection. Therefore we can say that - P(series) < p(parallel)


1

First you need to define the orientation of the cube relative to the axis you want to measure. Typically a 3×3 rotation matrix $E$ does the job transforming local coordinates along the principal axes to the world coordinates. The use the transformation $E I_{body} E^\intercal$ Example: A single rotation $\theta$ about the world $z$ axis is $$E = ...


1

The ball was in flight for four seconds: we can safely say that the ball reached maximum height at $t = 2$. (The gravitational pull is constant and there are no other forces acting, so the flight path is symmetrical). The ball was stationary at $t=2$ so its speed is $=0$ So now use the formula $v= u + at$, where $a$= acceleration, $t$= time, $u$= initial ...


1

Molecules are not that large. The density of air at 1 Atm and 20°C is $2.5\times 10^{25}$ m$^{-3}$. So, the average spacing, using your teacher's method, is about $(2.5\times 10^{25})^{-1/3} = 3.4$ nm. The radius of molecular nitrogen is 0.2 nm. So, the diameter is about 0.4 nm. The means that only about $(0.4/3.4)^3 = 0.0016$ of the volume of air is ...


1

When you calculate the "parallel" vector, you should not use the dot product of $a\cdot b$ but instead the normalized dot product $$\frac{a\cdot b}{|b|}$$ times the unit vector $b$. The projection of $a$ onto $b$ should always be independent of the length of $b$.


1

Let $I$ denote our integral: $$ I=\int{{e}^{x^2}(1+erfi(x))} dx $$ Using IBP: $$ u=(1+erfi(x))\quad dv={e}^{x^2}dx\\ du=\frac{2}{\sqrt{\pi}} {e}^{x^2}dx \quad v=erfi(x)\frac{\sqrt{\pi}}{2} $$ You get: $$ I=(1+erfi(x))erfi(x)\frac{\sqrt{\pi}}{2}-\int{erfi(x) {e}^{x^2}dx}\\ I=(1+erfi(x))erfi(x)\frac{\sqrt{\pi}}{2}-\int{erfi(x){e}^{x^2}-{e}^{x^2}dx} ...



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