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No, the square root of $100\,\mathrm s$ is $10\,\mathrm s^{1/2}$, not $10\,\mathrm s$. One $\mathrm s^{1/2}$ equals approximately $31.623\,\mathrm{ms}^{1/2}$. Just like a square meter is not the same unit as a meter (and a square kilometer is certainly not 1000 square meters), a $\sqrt{\text{second}}$ is not the same as a second. In the case of running ...

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As you asked for methods I decided to post an alternative. Here we use the trick of eliminating the inhomogeneity. First, consider only the equation for $v(t)=\dot{x}(t)$, $$\dot{v}=\mu v^2-g.$$ The inhomogeneity $-g$ prevents us from simple integration of the ODE, so let's make it disappear by defining $w(t)=v(t)-\gamma$, which leads to $$\dot{w}=\mu ... 2 Let l be length of slope. With out friction:- l=0.5(g\sin{\theta})t^2 With friction:- l=0.5(g\sin{\theta}-ug\cos{\theta})(2t)^2 if you divide these two u\cot{\theta}=\frac{3}{4} So u=\frac{3}{4}\tan{\theta} Assume θ=45 degree u=\frac{3}{4} 2 with out friction:- there will be no uN component as u=0 so only force is mgsinθ acceleration provided by this is gsinθ so considering the equation s=ut+1/2at^2 u=0(here u is initial velocity) so l=1/2(gsinθ)x(t^2) With friction:- friction causes an opposing force =uN(u=cofficient of friction) so umgcosθ resultant force on box is mgsinθ-umgcosθ ... 2 No, you can't add potentials like this, specifically because the charge redistributes. For an example, let M be a small sphere and N be a large disk. Let M have some charge. The potential distribution is the same as a point charge of that magnitude. If N has no net charge we know the potential distribution from it-zero everywhere. If we bring ... 2 No, clearly not. As an object falls, it "accelerates downward", meaning the velocity of the object as it travels downward increases, reaching it's maximum velocity just before hitting the ground. So if it takes the object, say, 15 seconds to hit the ground after being dropped, then it will travel the least distance in the first second, and the most distance ... 2 See this answer at Physics SE. In short, this attractive inverse squares problem is somewhat ill-defined in 1 dimension. If you look at your problem as limit case of Kepler problem with angular momentum M\to0, then you have the particle moving from its initial position to the singularity of the potential, and then bouncing off it (because the ellipse ... 2 A quick way to get it: You know the lengths of the two vectors and that  \ \vec{A} \bullet \vec{B} \  is given by  || \vec{A} || \cdot || \vec{B} || \ \cos \theta \  . The magnitude of  \ \vec{A} \ \times \ \vec{B} \  is  || \vec{A} || \cdot || \vec{B} || \ \sin \theta \ ,  so find  \ \cos \theta \  and use  \ \sin \theta \ = \ \sqrt{1 - ... 2 The first phase accelerates for some time so you can compute the distance travelled and the speed at the end of this phase. The speed is constant for the second phase and you know the duration. For the final phase, you know the speed and deceleration so you can compute the time until the train stops. This allows you to compute the distance taken to stop. ... 1 I will assume that these are the equations you need to use to solve this problem. To find the total distance covered is to find the sum of the distances it travels during acceleration at constant speed during deceleration Only three pieces of information about how the train accelerates are given: initial speed (0 m/s) acceleration ... 1 First, call its initial velocity v and height h. Here is the equation modelling the trajectory of the projectile: y = h + x \tan \theta - \frac{g}{2v^2} x^2 (1 + \tan^2 \theta). We want to find where the projectile hits the ground, so let y = 0. We can now use calculus to find the maximum by differentiating w.r.t \theta.$$x \sec^2 \theta + ...

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Note that the initial condition on $T$ is $T(x,0)=0$. But the equation you are really solving is for $U$, which has initial condition $U(x,0)=-x/a$. Thus, assuming you could derive $$U(x,t) = \sum_{n=1}^{\infty} B_n e^{-n^2 \pi^2 \kappa t/a^2} \sin{\left ( \frac{n \pi x}{a}\right )}$$ then $$U(x,0) = \sum_{n=1}^{\infty} B_n \sin{\left ( \frac{n \pi ... 1 If I were asked to prove the motion of the object as the equation given m \frac{dv}{dt} = -a|v| - mg Can i just do like this? Retarding force (a|v|) and gravitational force (mg) are the drags that pull the projectiled object.$$ F=-(a|v| + mg) \tag 1  F=ma \tag 2 $$so substracting (2) into (1):$$ ma=-(a|v| + mg) \\ m \frac{dv}{dt} ...

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Let $p(t)$ represent the vertical distance of the book from the ground at time $t$. We know that $p(0)=h$, where $h$ is the initial height from the ground, and that $p(0.430)=0$. Moreover, we know that $p''(t)=-9.8$ (the downward acceleration is constant). Can you think of where to go from here?

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The net force on the two boxes in contact is the vector sum of the applied ("pushing") force $\ F_P = 8.7 \$ N and the kinetic friction force $\ f_k \$ which acts in the opposite direction. The magnitude of that frictional force is 0.04 times the normal force of the force vertically upward on the boxes. Since the boxes are taken to be on a horizontal ...

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The paper, Use of the Projected Torus Knot Lattice for a Compact Storage Ring FEL, by S. Sasaki, A. Miyamoto such an example. In this paper the authors used the concepts of bundles and crossing numbers of several torus knots to determine how to design a free electron laser with required attenuation.

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You're mixing up the integrand and the integral. The knot's energy here is the integral of the energy density along the curve. Observing that the integrand blows up when $x=y$ is not sufficient to conclude that the integral is infinite; there are functions which blow up, yet can be integrated to a finite number. An example of such a function: $$... 1 To expand a bit on Henning's answer: the n that shows up in asymptotic analysis is generally not a dimensioned value at all - which is a good thing, because otherwise you wouldn't be able to (for instance) take its logarithm in a dimensionful fashion for a \Theta(n\log n) algorithm like, say, quicksort! Instead, n is a so-called dimensionless ... 1 Trigonometry was originally developed for solving problems in astronomy, and it took a long time before it was used for other purposes. Ptolemy's version of trigonometry was chord tables, later Indian astronomers or mathematicians found that the sine function was more convenient to use. Gauss developed the method of least squares to solve the problem of ... 1 Let's call the vertical cart Body 1 and the one moving horizontally Body 2. If the tension forces do not cause the string to break (and of course ignoring air resistance, friction, et cetera) then the vertical cart, Body 1, should accelerate downward at the same rate as the horizontal cart, Body 2, should accelerate to the right. If this wasn't the ... 1 I don't understand the overall picture, but here are some possible sources of problems: The cross product of a vector with itself is always the zero vector. So this line of code Vector3 crossProduct = Vector3.Cross(rateOfSpin, rateOfSpin); will give you \text{crossProduct} = (0,0,0). Then dotProduct and lengthOfDotProduct will also be zero. Then, ... 1 Think of it this way: You have the standard x- and y-axes centered on the car. For a flat road, the positive y-axis points in the direction of the normal force. When we tip the road, you have to tip the axes as well. Since we tip the negative x-axis up \beta, that tips the positive y-axis to the right \beta. The normal force is making an angle ... 1 To the problem with vectors, we have to break the trip up into two parts, I'll call them A (airport to change heading) and B (change heading to emergency landing) like you did. From the picture we can see that the vector A uses an angle of 68^{\circ} from the vertical. So A = \langle 140\sin(68^{\circ}), 140\cos(68^{\circ})\rangle \approx \langle ... 1 Use the fact$$ \ddot{x} = \dot{x}\frac{d}{dx}\dot{x} = \frac{d}{dx}\frac{\dot{x}^{2}}{2} $$then the equation becomes$$ \frac{d}{dx}\frac{v^{2}}{2} = \mu v^{2} - g $$Then you can solve directly to get$$ v^{2}\mathrm{e}^{-2\mu x} = \int 2g \mathrm{e}^{-2\mu x} dx + C_{1} = -\frac{g}{\mu}\mathrm{e}^{-2\mu x} + C_{1} $$if \dot{x}(0)=x(0)=0 then$$ ...

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I'm taking the angle of the box to the slope as $\theta$, which may be different from the angle you're using (but will pretty much give the same answer anyway if you adjust it). We can use $F=ma$, which gives us $a = g\sin{\theta} - \mu g \cos{\theta}$ in the case of friction. This we can plug into an equation for time, taking the length of the slope as an ...

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