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8

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 ...


3

I will not answer your question directly; only give you some tools that should help you answer the question (in practice) yourself. To focus the attention, find below a typical heating/cooling diagram for a frozen pure substance. The vertical axis marked $T$ represents temperature (in degrees Celsius). Three significant temperatures are indicated on the ...


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 ...


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

You can't "simply" calculate the average velocity from the velocity at the end points, unless the velocity graph is a straight line. Which it is between 2 and 3. But not between 0 and 3. So the approach you can take is this: What is the distance after 2 seconds of moving at 4 m/s? And what is the average velocity between t=2 and t=3 seconds (straight ...


1

The final position will be the initial position plus the area under the velocity versus time graph. That is the area between y =0 and the velocity function. I'm assuming you're not familiar with integral calculus, but if you look at the dimensions you arrive at by calculating this area you will find that it is meters. Good luck.


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} ...


1

As a simplification, you can consider that you have a 2D viscous flow between two boundaries that approach each other. Assuming that the flow is symmetrical about the line (with the line along the Y direction), you can simplify this further to "no flow at x=0". What you are left with is a pressure distribution $p(x,t)$ whose integral in $x$ should equal the ...


1

As @KyleKanos says, establishing a relationship with only two data points is dangerous. However, with that warning, if you want to test a relationship between quantities which you think might have some proportionality, you can take ratios of like quantities. Then test whether there is a relationship between various powers of the ratios. For example, if ...


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

"I don't know any equations..." is the point of dimensional analysis! Let's make a table of the quantities you listed, and their dimensions: M L T G -1 3 -2 \ c 1 -1 +- given these inputs... h 1 2 -1 / ------------- E 1 2 -2 - I need to get this output If we assume there is an expression $$E \propto G^A c^B h^C$$ then ...


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

Your work is fine (depending on your units) and what you were asked to show is wrong. Though I do object to saying you have a force equal to $ma_c,$ I would just say that a net force orthogonal to the velocity makes it go in a circle of radius $r$ where $F=mv^2/r.$ And the problem is famous. The fact that the frequency doesn't depend on the velocity or the ...


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 ...



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