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5

The acceleration of an object is the derivative of its velocity. The force acting on the object produces an acceleration according to the usual rule $F = ma$. The equation you get is $$1000 v'(t) = -200 v(t).$$ In other words, $v'(t) = - \frac 15 v(t)$ which has the general solution $v(t) = v(0)e^{-\frac 15 t}$. If $v(0) = 30$, you get $v(5) = ...


3

Oh wow! The answer just came to me! I'm not sure if this will help anyone, but I figure it's probably better to just provide the answer for posterity. So, really what we're talking about here is: $$\frac{dv}{dt} = \frac{-v}{5}$$ So, $$\frac{dv}{v} = \frac{-1}{5}dt$$ Then integrate both sides to get $$ln(v) = \frac{-1}{5}t + C$$ which rearranges to ...


2

Hint: Try to visualize the motion of the body separately along the 2 axes X and Y. Write the equations of motion with constant acceleration and you should have solvable equations for the height and velocity of the body. Try to think of the range of the body as the x component of the distance it travels when y increases from 0 and goes back to 0 as the ...


2

The radius $R$ of the Earth is about $3444$ nautical miles. Let $m$ be the height of the antenna (above sea level) in nautical miles. Let $O$ be the centre of the Earth, let $T$ be the top of the antenna, and let $H$ be a point on the horizon, as viewed from the top of the antenna. Let $d$ be the distance from $T$ to $H$. This is very close to the distance ...


2

Well, first you should recognize that we need some way of representing the $B$ you have in your first definition in the second definition. This is because the first definition is the definition of a linear mapping with the thing it maps (a bivector) and the second definition is just a linear mapping without the thing that it maps. In fact the second ...


2

I think the key here is to understand what is meant by the tensor product. For any vectors $a, b, u, v$, the tensor product $a \otimes b$ means $$(a \otimes b)(u,v) = (a \cdot u)(b \cdot v)$$ Or, perhaps, it might mean this instead: $$(a \otimes b)(v) = a (b \cdot v)$$ The two notions are equivalent to each other, so mathematicians freely use one or the ...


2

Here are a couple of books that aren't necessarily meant for mathematicians but are pretty mathematical: Foundations of Classical Electrodynamics: Charge, Flux, and Metric by Friedrich W. Hehl and Yuri N. Obukhov. This book is probably closest to what you're looking for. It is very rigorous, even starting by stating a set of axioms for classical ...


2

First make yourself clear what the intended statement is. For a neutron, the electromagnetic force is not stronger than gravity. Also, your formulas are for much too differnt situations: One is for the force acting upon a moving charge in a given magnetic field, the other is for a (resting or moving, doesnÄt matter) mass in the gravitational field near the ...


2

$$\ \int_{r_0}^{r}\sqrt{\frac{r_0r}{2GMr_0-2GMr+v_0^2r_0r}}dr=$$ $$\ =\sqrt{\frac{r_0}{v_0^2r_0-2GM}}\int_{r_0}^{r}\sqrt{\frac{r}{\frac{2GMr_0}{v_0^2r_0-2GM}+r}}dr$$ Where I assumed that $\ v_0^2r_0>2GM$. Now set $$\ \beta=\sqrt{\frac{r_0}{v_0^2r_0-2GM}}$$ $$\ \alpha=\frac{2GMr_0}{v_0^2r_0-2GM}$$ So you need to deal with this kind of integral: $$\ ...


2

If three vectors are in equilibrium, then there is one very important relationship between these vectors: that $\mathbf{A} + \mathbf{B} + \mathbf{C} = 0$. Hence, $\mathbf{A} \times (\mathbf{A} + \mathbf{B} + \mathbf{C}) = \mathbf{A} \times \mathbf{A} + \mathbf{A} \times \mathbf{B} + \mathbf{A} \times \mathbf{C} = \mathbf{0} $ This means, $\mathbf{A} ...


2

You can't multiply by $\epsilon^{ijk}$ because there's already a $k$ in $B_k$ and indices should not appear more than twice.


1

It is easy, you can write: $\alpha(t)=(1-t)(0,1)+t(2,4)=(2t, 1+3t)$ with $t\in [0,1]$. And therefore: $$dW=F(\alpha(t))\cdot \alpha'(t)dt$$


1

The solution is no more complex. You have to use the equation $y(t)=h_0+v_0t-\frac{1}{2}gt^2$, where $h_0$ is the initial position ($40$m), and $v_0$ the initial velocity ($0$ in this case). Now, solve $40-4.9t^2=0$ and use the original fórmula for the other questions too.


1

Hint: You have a problem in two dimension with: $ \vec v_0=(v_0,0)^T$ (and I suppose your velocity is in meter/sec) $\vec a=(0,-g)^T$ $\vec s_0=(40,0)$ Your equation of motion become: $$ \vec s(t)=\dfrac{t^2}{2}\vec a+t \vec v_0+\vec s_0 \Rightarrow \begin{bmatrix}x(t)\\y(t) \end {bmatrix}=\dfrac{t^2}{2}\begin{bmatrix}0\\-g \end ...


1

Don't feel bad.I think the wording of the problem is incorrect.If I had to guess,it would go something like this:"A stone is dropped vertically from a height of 40 meters above the ground at an initial velocity of 15 meters/sec. We need to measure distance from 40 meters up toward the ground,If s = distance, s = 0 at 40 meters up.The acceleration of gravity ...


1

Firstly you shouldn't use $x$ for your scientific notation --- use $\times$ (\times). Also a good convention with units is to leave spaces between the number and the unit (and other units)... for example $20\,\text{m s}^{-1}$ is better than $20\text{ms}^{-1}$. So anyway we have a density of $\rho=7.86\times 10^3$ kg m$^{-3}$. Note we define density as ...


1

Two first places to look, and a difficult option: Dimock, Quantum Mechanics and Quantum Field Theory. Covers the mathematics behind both of the named topics. This was the book my supervisor recommended to me. Goes through fairly axiomatically, first non-relativistic, then QFT, then some stochastic stuff. Has an appendix of the functional analysis you need. ...


1

$dr$ is the infinitesimal change in the radial direction while $r d\theta$ is the corresponding change in the perpendicular direction. The arc length is therefore: $$ ds = \sqrt{dr^2 + (rd\theta)^2} = \\ \sqrt{1 + \left(r\frac{d\theta}{dr}\right)^2}dr $$ Also, $tan\beta$ is the ratio of tangential to the radial displacement i.e. $$ tan\beta = ...


1

The answer $144g\ J$ is for convenience and exactness. Note that $144 \times 9.81 = 1412.64\ $ in correspondence with the magnitude of your answer. The units are okay, if we read between the lines. Your professor was avoiding clunky notation: $(144 kg\cdot m)g$ where $g = 9.81 m/s^2$. It's just more convenient to "strip g of its units" and state that the ...


1

The problem posed is the first (and easy) part of a tougher problem. You start with a string which is flat at $t=0$ and is not moving at that time. Then instataneously one side of the string is raised to $H$, producing a delta-function pulse in the string, and thereafter the ends are held at $0$ and $H$ and the tension is such that the wave-speed is $c$. ...


1

Here is how you go about solving this question. Since the snowballs are being thrown at angles, you need to separate your initial speed into $i$ and $j$ components. So for the case of Snowball1: $v=12m/s$ at $40$ degrees $u_i = 12\cos(40)$ $u_j = 12\sin(40)$ The formula you need is $v^2 = u^2 +2as$ First, find final speed in the $j$ direction. List ...


1

Consider getting to the midpoint of the trip at $X/2$ in time $T/2$. Thus $$ \frac{X}{2} = \frac12 g \left( \frac{T}{2} \right) ^2$$ You can solve this for $T$ in terms of $X$ and $g$: $$ \frac{g}{8}T^2 = \frac{X}{2} \\ T^2 = \frac{4X}{g} \\ T = 2\sqrt{\frac{X}{g}} $$


1

In the simplest model, you are working in two dimensions, with the system of equations: $$\frac{d^2 y}{dt^2} = -g \\ \frac{d^2 x}{dt^2} = 0 \\ y(0)=0,y'(0)=\sin(\theta) s \\ x(0)=0,x'(0)=\cos(\theta) s $$ where $g>0$ is acceleration due to gravity, $\theta$ is the angle of the initial velocity, and $s$ is the initial speed. Note that this model assumes ...


1

In projectile motion, you need to concern yourself with the two components of displacement: $$x=x_0 + v_x t + \frac{1}{2}a_x t^2$$ $$y=y_0 + v_y t + \frac{1}{2}a_y t^2$$ If we place the cannon at the origin, then $(x_0,y_0) = (0,0)$. Furthermore, there is no acceleration in the $x$ direction if we neglect things like air resistance. Then $a_x = 0$. The ...


1

$\alpha$ represents the phase. You are correct that if you include $\alpha$ you can use either $\cos$ or $\sin$ to represent the motion-it will just shift $\alpha$ by $\frac \pi 2$. You can expand your sine wave into $y(t)=A\sin(\omega t) \cos (\alpha) + A\cos(\omega t)\sin \alpha) +B$ Substituting in $t=0$ shows that $y(0)=A\sin (\alpha)+B, y'(0)=A ...


1

Let denote: $$ s_k=\sum_{i=1}^k a_i\ \ \ \ \ t_k=\sum_{i=1}^kb_i$$ hence the inequality to prove is : $$P(n)\ \ \ \ \ \ \ \ \ \ 1+\frac{t_n}{s_n}\leq \prod_{k=1}^{n} \frac{s_k+t_k}{s_k+t_{k-1}}=\prod_{k=1}^{n} \left(1+\frac{b_k}{s_k+t_{k-1}}\right)$$ Proof (by induction on $n$) Basis step $P(1)$ is obviously true because: $$(1+\frac{b}{a})\leq ...


1

This quote from American Practical Navigator (Bowditch, 1981) Volume II, Article 506, may be useful: "Radar horizon. The distance to the radar horizon is the distance between the transmitter and the point at which the radar rays graze the surface of the earth. In the standard atmosphere, radar rays, like light rays, are bent or refracted slightly ...


1

This is the well-known derivation of the kinetic energy formula. You'll find it easier to work in scalars initially to see what's happening - so let's make the assumption that the force is always in the direction of motion (thereby obviating the need for the dot products). The derivation is a "shortcut" application of a change of variables and uses chain ...


1

Recall the definition of $a$: $a = \dfrac{dv}{dt}; \tag{1}$ thus, with the numbers provided, $\dfrac{dv}{dt} = -\dfrac{1}{5}v; \tag{2}$ this yields $v(t) = v(t_0)e^{-(t - t_0) / 5}; \tag{3}$ here I have supplied the number not explicitly provided, which is $v(t_0)$, the initial velocity at time $t = t_0$; apparently $v(t_0) = 30 M/sec$, assuming we ...



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