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8

Maybe this helps: link Looks like some time ago the second was defined by $1/2$ of the oscillation time of a $1$ meter long pendulum. The oscillation time of a pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$. With $T = 2$ and $L = 1$ this gives $g = \pi^2$


4

According to your information, air resistance $\mathbf{f} \propto -v^{2}$ and \begin{align*} \frac{\mathbf{f}}{m} &= -bv^{2} \hat{\mathbf{v}} \\ &= -bv^{2} \frac{v_{x}\, \mathbf{i}+v_{y}\, \mathbf{j}} {|v|} \\ &= -b|v|(v_{x}\, \mathbf{i}+v_{y}\, \mathbf{j}) \\ ...


4

You are right, the partial derivatives of distributions on higher-dimensional spaces are defined as you surmised, $$\frac{\partial T}{\partial x_i} \colon \varphi \mapsto -T\biggl[\frac{\partial \varphi}{\partial x_i}\biggr],$$ more generally $$D^{\alpha} T \colon \varphi \mapsto (-1)^{\lvert\alpha\rvert} T[D^{\alpha}\varphi]$$ for higher derivatives. ...


3

$\mathbf{b-a}$ is a vector in the direction from $\mathbf{a}$ to $\mathbf{b}$. Dividing by its length produces a unit vector in that direction. Then multiplying by the (presumably positive) scalar $u$ produces a vector of magnitude $u$ in the direction from $\mathbf{a}$ to $\mathbf{b}$, which is precisely what you want as the velocity vector.


2

You need a unit vector for the direction of motion, otherwise the speed gets multiplied by the length of the vector, so the net speed is off. Dividing $\mathbf b-\mathbf a$ by its length takes care of that.


2

Inspired by Frentos's answer, for the time being, let us assume something like $$y=\frac ax+b$$ Using your values, a simple linear regression gives $$y=\frac{0.541976}{x}-0.000038024$$ which gives as predicted corrections $0.01803$, $0.00538$, $0.00267$, $0.00177$. As said, with more data points, something better could come out. Edit Take care; the ...


2

And the resulting would be your Cross Product or the coordinates of an orthogonal vector. My question is why? Why does forming it that way give you the magnitude of an orthogonal vector Your last equation can be written as $$ (a \times b)_i = \epsilon_{ijk} a_j b_k \quad (1) $$ where $\epsilon_{ijk}$ is the skew-symmetric or Levi-Civita tensor and ...


2

The difference between the way that Mathematicians and Physicists handle such things is based in an underlying assumption of Physics that solutions of Physical equations exist. The Mathematician would say that, if you have a $C^2$ scalar field $F$ in a region of space, then you can reconstruct $F$ if you know $\nabla^2F$. First, for $r \ne r'$, $$ ...


2

Hint : $$a=\frac{v-u}{t}, at=v-u$$ So what is $ut+\frac{1}{2}(v-u)t?$


2

By area under graph we find (add your rectangular and trianguar components not multiply) that: $$s=ut+\frac{t(v-u)}{2}$$ now $a=\frac{v-u}{t}$ so multiplying both sides by $t^2/2$ gives $\frac{at^2}{2}=\frac{t(v-u)}{2}$ and the answer follows.


2

Consider a particle which passes through a particular point going in one direction, then loops around and passes through the same point again going in the opposite direction. Then at some particular position, the particle has two different velocities. So the velocity of this particle is not a function of its position, in the sense that if $v(t)$ denotes the ...


2

First, we establish a simple result. Suppose we have two operators $A,B$ with $[A,B]=1$. Then one can easily prove via induction that more generally $[A,B^k]=k B^{k-1}$ for any natural $k$. Therefore the map $d_B$ defined as $d_B(\cdot)=[A,\cdot]$ acts as a derivative on all such terms $B^k$, and since this map is linear we furthermore have that this extends ...


2

I agree with you that one good reason for using $\cos$ is that it corresponds well with the initial conditions for harmonic motion. Another good reason is experimental observation. If I come across a physical system that is oscillating, I will want to measure its amplitude and period. Examples include a swinging pendulum, a planet observed rotating around ...


2

In a topological space $X$, a set $A\subset X$ is dense if $\forall x\in X$ any neighborhood of $x$ contains at least one point of $A$. In the physical sense: given a specified volume and a mass: $$ \text{density}=\frac{\text{mass}}{\text{volume}}.$$ We do not have a notion of volume/distance in a general topological space, nor do we have a notion of ...


2

The problem is that you didn't specify what "mathematical space" or "dimension" or "axis" means at all, so there is no answer to your question. For example, I could just map every object $x$ in some model to the pair $(x,0)$, and modify all functions and predicates, and then indeed the resulting collection of objects is isomorphic to the original. So what? ...


1

Another reason is to use the initial position as a parameter. Compare $z=z_0\lambda^t$ and $h=h_0\cos \omega t$ You can't do the same with $\sin$ - at best you can say perhaps $d=d_{max}\sin \omega t$ By the way, this argument fails once you introduce a phase shift!


1

Assuming your computation of the fourth force being $(-87.34,-22.54)$ is correct, then what you want is $75.5^\circ+180^\circ$. It may help to note that $\arctan(-87.34/(-22.53))=\arctan(87.34/22.53)$, which will give you one of the angles of a right triangle with legs $87.34$ and $22.53$. However, your fourth force lies in the 3rd quadrant, so you should ...


1

We have \begin{align} \text{average velocity}&=\frac{x(4.1)-x(0)}{4.1-0\text{ s}}\\ &=\frac{(1.45\text{ m/s}^2)(4.1\text{ s})^2-(0.055\text{ m/s}^3)(4.1\text{ s})^3-0}{4.1\text{ s}}\\ &\approx \color{blue}{5.02\text{ m/s}} \end{align}


1

Unless you have a picture or anything else to indicate horizontal motion, I'd take it that the motion occurs only up-and-down and that the (upward) launch from the cliff is just a way of getting an initial non-zero displacement. And presumably the cliff is perfectly vertical and the projectile motion takes place an infinitesimal distance away from the cliff ...


1

When you compute the work in your integral, the force should not be the force of gravity on the whole tank, but just on a slice of thickness $dh$, so the force should be $g\rho \pi r^2 dh$. That is because you pump that slice up a certain distance and other slices up a different distance. For part B, you can just lift all the water $2$ meters, then use the ...


1

For part (a) we're assuming pumping puts all the water at the same level. So water at a lower level requires more work to pump out. Imagine a cylinder of water at depth $x$ meter (measured from the top of the side of the pool) of height $dx$ meter in the pool. Mass of this cylinder = $49000\pi dx$ kg. Work needed to pump this cylinder out is ...


1

Toss a fair coin. For a single drawing the expected proportion of heads is distributed as follows (for $0$ and $1$): $$\frac12, \frac12$$ For ten drawings in a row, by simple probability computation (from $0$ to $10$): $$\frac{1}{1024}, \frac{10}{1024}, \frac{45}{1024}, \frac{120}{1024}, \frac{210}{1024}, \frac{252}{1024}, \frac{210}{1024}, ...


1

I hope I have a possible solution. Le us assume that, for all the components $J_i$ of $\boldsymbol{J}$, there is a function $f_i\in C^4(\mathring{A})$ such that $\nabla^2 f_i=J_i$, with $\bar{V}\subset \mathring{A}$ and $\boldsymbol{x}\in \mathring{V}$ and there exists a $\delta$ such that, for all $\epsilon\le \delta$, $\epsilon >0$, the region ...


1

Another way is to notice that as the speed is a linear function of time, the average speed is the speed at half the time (also the average of the initial and final speeds): $$\bar v=u+a\frac t2,$$ so that the space is $$s=\bar vt=ut+a\frac{t^2}2.$$


1

The amount of the sphere at a distance of $r$ is $$ 4\pi r^2\,\mathrm{d}r $$ so the average of $r^2$ would be $$ \begin{align} \frac{\int_0^{r_0}4\pi r^2r^2\,\mathrm{d}r}{\int_0^{r_0}4\pi r^2\,\mathrm{d}r} &=\frac{\frac{4\pi}5r_0^5}{\frac{4\pi}3r_0^3}\\[3pt] &=\frac35r_0^2 \end{align} $$


1

Looking at the computation from the right angle, what you compute is a new vector in the dual space, such that $\vec a \times \vec b$ maps any vector $\vec c$ to $\mathbb R$ (or $\mathbb C$), in a way that $\det(\vec c, \vec a,\vec b)=(\vec a\times\vec b)\cdot \vec c$. You can see this by replacing $(i,j,k)$ by $(c_1, c_2, c_3)$. Everything else follows ...


1

If you start with the definition of cross-product as $$\underline{a}\times\underline{b}=|\underline{a}||\underline{b}|\sin \theta \underline{\hat{n}},$$ where $\theta$ is the angle between $\underline{a}$ and $\underline{b}$ and $\underline{\hat{n}}$ is the unit vector perpendicular to $\underline{a}$ and $\underline{b}$ in the sense of a right-hand triad' ...


1

The determinant of a $3\times3$ matrix can be viewed as the triple product of its columns (or rows): $$ \begin{align} \det\begin{bmatrix} x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3 \end{bmatrix} &= \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} \times \begin{bmatrix} y_1\\ y_2\\ y_3 \end{bmatrix} \cdot \begin{bmatrix} z_1\\ z_2\\ z_3 ...


1

For a fixed $x_0$, define $G_{x_0}=|x-x_0|^{-1}$. Then $\nabla^2 G_{x_0}=0$ for $x \ne x_0$. If $\varphi$ is a compactly supported $C^{\infty}$ function on $\mathbb{R}^3$, then $$ \nabla\cdot(G_{x_0}\nabla\varphi-\varphi \nabla G_{x_0})=G_{x_0}\nabla^2\varphi-\varphi\nabla^2G_{x_0}=G_{x_0}\nabla^2\varphi,\;\;\; x \ne x_0. $$ Integrate and apply the ...


1

to get $q$ just plug in the values without the $\pm$ $$q = \frac{x^2 + ysin\theta + 2}{x + ycos4\theta}$$ for the uncertainty $\Delta q$ , first convert $\theta$ and $\Delta \theta$ from degrees to radians and then use the formula ... $$ \Delta q = \sqrt{ \left( \frac{\partial q}{\partial x} \Delta x \right)^2 +\left( \frac{\partial q}{\partial y} ...



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