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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 (or sometimes below) 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 ...


3

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 ...


2

The simplest way is to exploit the symmetries here. So, instead of mindlessly going through the algebra, solving equations etc. you just use the fact that you are free to call any direction the x-direction, and set up a right handed coordinate system. In particular, this means that you are free to cyclically permute x, y, z in the equations. The problem of ...


2

Take the kinematic relationship $$ x_{end} - x_{start} = \frac{ v_{end}^2 - v_{start}^2 }{2 a }$$ which applies to constant acceleration and use it twice. First on the first interval to calculation the acceleration $a$ and then again on the second interval to find the distance traveled.


1

Yes, you should start by assuming the wavelength used, $\lambda$, as known. The information you have is just sufficient to extract $\lambda$ in terms of the relative shift $\eta = \Delta \lambda / \lambda$ and the scattering angle $\theta$. All you have to do is maximize $\lambda$ in terms of $\eta$, $\theta$.


1

There are four kinematic equations that apply to this type of problem. One of those equations can be used to solve for acceleration when you don't how long it took for the car to stop. The equation is: $v_f^2 = v_i^2 + 2a \Delta\ x $ The initial velocity is given, the final velocity is zero, and the distance traveled is given. The only unknown, ...


1

Your moment of inertia is incorrect. You must calculate is based on the individual masses and their distances from the pivot: $$\mathcal{I}=\Large\Sigma \large\left( m_ir_i^2\right).$$ If you do this you should get an answer that agrees with what @ChrisDrost did, 2.47 s And you shouldn't assume that the center of mass is 1/3 of the way below the rotation ...


1

For error propagation for one variable, it is best to use $$\delta f(x) = \left|\frac{d f(x)}{d x}\right| \delta x$$ which is to say that the uncertainty in the function should be weighted with the derivative(or how sensitive the function is to changing the variable) Now for your example, $$f(x) = x^n \Rightarrow \delta f(x) = \left|n x^{n-1} \right| ...


1

Steve B is right. I just want to include a diagram. You can see that the surface $\rm S$ consisting of the conducting surfaces $\rm S_1, S_2, S_3,...., S_i$ enclosing the volume $\rm V$. Suppose $\rm S_1$ is at unit potential and all the others are at zero potential. If $\rm S_1$ has $\rm Q$ amount of charge then an equal amount of negative charge is ...


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

By simple calculus: Between $t=2s$ and $t=4s$, $v_x$ can be described by the function: $v_x=8-2t$. The definition of $v_x$ is: $v_x=\frac{dx}{dt}$, so that $dx=v_xdt$. Integrating between $t=2$ and $t=3$ we get the distance travelled in that interval: $x=\int_2^3(8-2t)dt=3m$. Add to this the $8m$ travelled in the first $2s$, so total distance is ...


1

Let me first do this the way that I know is correct: with Lagrangian mechanics. This says that all of the physics you need is contained in the Lagrangian, which is the kinetic energy minus the potential energy. Your three masses Left, Right, and Bottom make the kinetic energy $K = \frac 12 m (v_L^2 + v_R^2 + v_B^2),$ where $m = \text{1 kg}.$ Defining ...


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

You've almost got it! The constant thrust comes from a mass rate $\mu$ of fuel being expelled at a velocity $v_f$ as opposed to the speed of the rocket itself $v$. Therefore the equation is instead: $$ (m_0 - \mu t) \frac{dv}{dt} = \mu v_f - \alpha \frac {(m_0 - \mu t)}{r^2},$$ where $\alpha = G M.$ Hence the gravitational term you wrote as $G m_e (W_0 + ...


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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