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Note: The solution below is long only because I'm being very detailed, for pedagogical reasons. The net force acting on the package is the vector sum of the force of gravity and the wind resistance: $$\mathbf{F}_{\mbox{net}} = m\mathbf{g} - \alpha\mathbf{v}$$ Newton's second law of motion then tells us that the acceleration of the package, ...

4

$A$ and $B$ are two points on the same side of the mirror. A ray of light from $A$ reflects off the mirror and goes to $B$. If I understand correctly, what is to be proved is this: among all paths going from $A$ to the mirror and then to $B$, the shortest one is the one for which the two angles are equal. A way to prove that is to consider the ray of light ...

3

The second (worse) version of your algorithm resembles the explicit Euler scheme. It is consistent to first order only, and it does not preserve invariants of motion such as the total energy. In fact, the errors of the Euler scheme act as excitations, so the orbits of your planets would spiral outward over time. Runge-Kutta methods are designed to be ...

3

Integrating once (using $x'(0)=0$), from zero to $\tau$, we have $$x'(\tau)=\int_0^\tau F(s)\,ds-\tau.$$ Integrating once more (using $x(0)=0$), from zero to $t$, we find $$x(t)=\int_0^t \int_0^\tau F(s)\,ds\,d\tau-t^2/2.$$ In the integral, we integrate by parts, $$\int_0^t 1 \int_0^\tau F(s)\,ds\,d\tau=\Bigl[(\tau-t)\int_0^\tau ... 3 Let's first make a free body diagram: As you said, when the speed is maximum the acceleration is zero, and since the motion is only along the plane and never leaves it, we will be only dealing with the horizontal components:$$\begin{align} P_x+N_x+f_x+F_x&=0\\ -mg\sin(\alpha)-f+F&=0\\ F&=f+mg\sin(\alpha)\tag1 \end{align}$$F is constant, ... 2 Maybe you started by asking for something like this... ASSUMPTION. We start at time t = 0 and consider values t \ge 0. We end when the rocket is on the ground, that means height 0, so we must have h(0) = 0. We need to find t such that h(t) = 0. GIVEN. h(t) = 160 - 16t^2. The rocket must start at its highest point because h(0) = 160 and the ... 2 Runge-Kutta is a typical choice. Also, I would highly recommend that you take a look at Introduction to Computer Simulation Methods. The book covers a wide range of topics relating to simulating physical phenomena, from solar systems to molecular diffusion to galaxies to chemical reactions, and more. I personally like the 2nd edition better than the 3rd ... 2 I presume your integral is over all R^N and \nabla u=\partial u/\partial x is a vector in R^N. Then:$$ \begin{aligned} T(u_\sigma) &= \int_{R^N} |\nabla u_\sigma(x)|^2\, dx = \int_{R^N} |\nabla u(x/\sigma)|^2\, dx = \int_{R^N} \left|{\partial u(x/\sigma)\over\partial x}\right|^2\, \sigma^N d(x/\sigma)\\ &= \int_{R^N} \left|{\partial ...

1

This will be a 2-dimensional graph, since the position vector r has 2 components. Draw an x,y plane. At time t, the x position is $\cos(\omega t)$ and the y position is $\sin(3\omega t)$.

1

The main source of confusion here is the definition of a "pound". You say the mass of water that strikes the plate each second is $$(\text{volume of water per second}) \times 62.5\ \text{pounds}.$$ But actually that is the weight of the water. The mass of the water, which you need in order to correctly compute $\frac{\Delta mv}{t}$, is the $m$ in the ...

1

Notice, the force exerted on the fixed plate in the direction of the jet line$$=\text{rate of change of momentum of water striking the plate}$$$$=\text{mass/sec} \times\text{(velocity before striking-velocity after striking)}=(\rho a V)(V-0)=\rho aV^2$$ Where, density of water, $\rho=1000\ kg/m^3$ Speed of water jet, $V=50\times \frac{30.48}{100}=15.24\ ... 1 Like every statement in natural language, every mathematical statement does not occur isolated but in a certain context. There are also undefinable concepts in any language, whether natural or not. "$1+1=2$" is only meaningful in a context where "$1$" and "$2$" and "$+$" and "$=$" are meaningful. If their meanings are completely determined, then that ... 1 AFAIK, there's no reason to think that mathematics would be different in any other universe, nor that it could be different. 1 A few things get swept under the rug in these formal manipulations. The first thing is that, as you've mentioned, these expressions should be read as $$\frac{\partial S}{\partial E}=\frac{1}{T} \\ \frac{\partial S}{\partial V}=\frac{p}{T}$$ where we hold$V$and$E$constant in the respective lines. An important fact here is the inverse function theorem, ... 1 For ease of understanding; let's assume$r(t) = at^2 + bt + cto be the radial change wrt time. You'll get the integral to be: \begin{align} I &= \int_{r_1}^{r_2}{2 \pi \cdot r \cdot h \cdot dr} \\ &= 2\pi h \int_{t_1}^{t_2}{(at^2 + bt + c)\left[(2at + b)dt\right]} \end{align} In other words: $$\dfrac{dr}{dt} = 2at + b \\ \equiv dr = (2at + ... 1 When the rope is pulled beyond its natural length, the force it exerts (by Hooke's law) is given by F=kx, where x is the distance beyond the natural length that the rope is being stretched and k is a constant particular to the rope. A person of mass m will hang stationary on the rope if the force from the rope pulling them up, kx, is equal to the ... 1 As Rahul wrote, "a collision against a vertex or edge" seems to be intended to mean "any collision against a vertex or edge". This interpretation fits well with the rest of the text; in particular it makes sense of the immediately following sentence. 1 Contrary to my initial gut feeling, this sounds highly plausible. If you assume that any body which is not moving (with constant speed) along a geodesic experiences some force, then you can simply take four test bodies, forming a regular hyperbolic tetrahedron, and consider translating that through space. As you said, a hyperbolic translation only moves ... 1 You want to compute$$ \frac{n_1}{s_0} + \frac{n_2}{s_i} $$to the lowest non-vanishing power of h/R, given that$$ \frac{n_1}{l_0} + \frac{n_2}{l_i} = \frac{1}{R}\left( \frac{n_2\,s_i}{l_i} - \frac{n_1\,s_0}{l_0} \right) \qquad (1) $$where$$ l_0^2 = R^2 + (s_0 + R)^2 - 2R\,(s_0 + R)\cos\phi  l_i^2 = R^2 + (s_i - R)^2 + 2R\,(s_i - R)\cos\phi ... 1 You are on the right path. The driving force is\frac{\text{Power}}{\text{speed}}=3600g$Taking$g=10$, you get the expected answer. 1 We know that the value of Earth's gravitational acceleration on the surface is given as $$g=\frac{Gm_e}{R_e^2}=9.81$$ Where,$M_e=\text{mass of Earth}=6\times 10^{24}\ kg$&$R_e=\text{radius of Earth}=6400\ km$Consider an object of wight$W$on the Earth's surface Then, the weight of the object at a height$h$from the Earth's surface is given by ... 1 OK, you know the gravitational law: $$F = \frac{G M m}{r^2}$$ where$M$is the mass of the earth and$m$is your mass. You weigh, on the surface of the earth $$F_0 = \frac{G M m}{R_e^2}$$ where$R_e$is the radius of the earth. Your weight at a height$h$above is $$F_h = \frac{G M m}{(R_e+h)^2}$$ Can you now compare these weights? 1 METHOD 1: We can simplify the problem by noting that$\nabla \times \vec F=0$. Therefore,$\vec F$is therefore conservative on any connected domain. By Stokes' Theorem, the integral of$\vec F$over any contour$C$that bounds a connected domain$S$is $$\oint_C \vec F\cdot d\vec \ell =\int_S \nabla \times \vec F\cdot \hat ndS$$ Thus, the integral of ... 1 Here is a hint to get you started: Since$x(0)=0$,$x(t)=x(t)-x(0)=\int_0^t v(t)dt\;\;$where$v(t)=\begin{cases}\frac{5}{3}t &\mbox{, if } 0\le t\le3\\5 &\mbox{, if }3\le t\le 7\\-\frac{5}{3}t+\frac{50}{3} &\mbox{, if } 7\le t\le10\end{cases}$1 Recall that a potential function of$\mathbf{\vec F}$is a function$f$such that$\mathbf{\vec F} = \vec \nabla f$. We recognize that$\frac{\partial U}{\partial y} = \frac{x^2}{2}-yx^2+C'(y) \neq yx^2$for any choice of$C(y)$, so your answer is unfortunately incorrect. There is a small mistake in the third line of your computations. We have $$f(x,y) = ... 1 By the Taylor series expansion near 0, say for |u|<1, we have$$ \frac{1}{(1-u)^2}=1+2u+O(u^3) \tag1  \frac{1}{(1+u)^2}=1-2u+O(u^3) \tag2 $$giving$$ \frac{1}{(1-u)^2}-\frac{1}{(1+u)^2}=4u+O(u^3) \tag3 $$then, setting u:=\dfrac{l}x, we get from (3),$$ ... 1 There is no need to use Taylor series, just algebra and limits: $$H=\frac{M}{2l}\left(\frac{1}{(x-l)^2}-\frac{1}{(x+l)^2} \right) =\frac{M}{2l}\frac{(x+l)^2-(x-l)^2}{(x^2-l^2)^2} =\frac{M}{2l}\frac{4xl}{x^4(1-(l/x)^2)^2} \approx \frac{2M}{x^3}$$ 1 The mistake in the calculation is a sign error in your formula for the equilibrium temperature$T$. The heat lost (or gained) by substance$i$is$m_ic_i(T-T_i)$. By the first law of thermodynamics we have $$m_1c_1(T-T_1) + m_2c_2(T-T_2) = 0 \implies T = \frac{m_1c_1T_1\color{red}{+}m_2c_2T_2}{m_1c_1\color{red}{+}m_2c_2}$$ As a consistency check we see ... 1 To reader: This answer addresses the inequality originally posted. This inequality doesn't hold. For$T_1 = T_2$we have: $$\Bigg(\frac{m_1 c_{p1}-m_2 c_{p2}\frac{T_2}{T_1}}{m_1 c_{p1}-m_2 c_{p2}}\Bigg)^{m_1 c_{p1}}\Bigg(\frac{m_1 c_{p1}\frac{T_1}{T_2}-m_2 c_{p2}}{m_1 c_{p1}-m_2 c_{p2}}\Bigg)^{m_2 c_{p2}} = 1^{m_1 c_{p1}}1^{m_2 c_{p2}} = 1$$ But there are ... 1 My confusion stemmed from mixing up$x$and$\theta$as well as not remembering to take$Df(x)$as$x = x_e = 0$. Worked out: From the expression for$\dot{x}$it is clear that $$f([x_1 \ x_2]^T) = \left[ \begin{array}{c} x_2 \\ -\frac{g}{\ell}\sin x_1 \end{array} \right]$$ so$\$\begin{align*} Df(x_e) &= \left[ \begin{array}{cc} \frac{\partial ...

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