Tag Info

Hot answers tagged

4

The speed of projection is minimal when the projectile just grazes the mound, or equivalently when the projectile fired tangentially to the mound in a vertical plane hits the apex of the mound. Taking the projection point as $(-a\sin\theta,a\cos\theta)$, with the $y$ axis vertical through the centre of the mound, the trajectory in this plane is given by ...


3

For concreteness sake, let's suppose we are computing the moment of inertia of the Koch Snowflake $K$ of uniform density and mass 1 shown on the left below. This snowflake is centered at the origin and has diameter 3 - that is, the maximum distance between two points is three. From here, it would be easy to compute the moment of another Koch flake of ...


2

Lets take the approximation $$ T = 2\pi \sqrt{\frac{L}{g}} $$ for the moon we have to compute $g(\text{moon})$ $$ g(\text{moon}) = \frac{GM_{\text{moon}}}{r^2} $$ this is the magnitude of the acceleration. $$ M_{\text{moon}} = \rho V_{\text{moon}} = \rho \frac{4}{3}\pi r^3 $$ now since we know that density ($\rho$) is consistent between earth and moon we ...


2

Almost all of it is correct, only the last equation is wrong: The ground is 30 meters UNDER the bridge, so you want the time at which the value od $s(t)$ is equal to $-30$. It will be very helpful for you to understand why I was able to very quickly notice that something is wrong with your solution. Here is the thought train that got me there: OK, ...


1

Don't know if this is the best way, but here goes: Your inertia matrix is real and symmetric. Therefore its eigenvalues must be real, and the determinant $\Delta = \omega_1 \omega_2 \omega_3$, the product of your eigenvalues. So find the eigenvalues of your matrix and you will find that none of them is 0, thus $\Delta$ will also be non-zero.


1

I'm gonna put in a long answer here since I forgot all of my kinematic equations (sad I know). From what I remember from taking a course on fundamental physics in college, acceleration is just the second derivative of displacement with respect to time or the first derivative of velocity $\vec a=\frac{d^2 \vec s}{dt^2} = \frac {dv}{dt}$ where $s$ is the ...


1

The bounds of the integral are being passed to (12-y), so smaller y results in larger areas of integration. The first equation seems easier to "visualize" to me too, but I think the second one can be intuited as well. The first term is the work done for the elevator. The second term is the work done for the cable as it moves from position zero to position ...


1

https://en.wikipedia.org/wiki/Cross-correlation is what you are looking for. If you had data points for 2 curves, you'd normalize them (make their sum equal 1) and perform "cross correlation" which is basically convolution without flipping anything. The maximum value (out of 1) will tell you how similar the two curves are and the position of the peak will ...


1

Here's the approach I would take. (1) Moments of inertia are additive, so the idea would be to do a summation over all iterations of triangle-adding. (2) The moment of inertia of the central triangle is straightforward to compute. (3) At the $i_{th}$ iteration, there are $a_i$ triangles. (4) The center of these triangles are a computable distance $d_i$ ...


1

Think of it like this, We start with the velocity being $-10$ (I.e. Going in reverse), and we accelerate at a constant rate of $2$. Eventually you reach a point where your velocity is $0$, and we start going forward after.


1

The key fact that makes this possible is the fact that position, velocity and acceleration are all vector quantities. Even if the magnitudes of these quantities is zero, the direction in a given coordinate system is not zero. Abel hints at this in his comment above-when a ball is thrown up in the air and reaches it's maximum height, the velocity is 0, but ...


1

Since the question is asking for distance travelled, the trick is to bring distance along the circular track, say $x$, in as the variable instead of time. So if we take $F = m \dfrac{dv}{dt}$, then note that $F\dfrac{dx}{dt} = mv \dfrac{dv}{dt}$. Now we can remove the explicit dependence on $t$ and obtain $$\int_0^{X} F\,dx = \int_{v_0}^0 mv\,dv.$$


1

Since the curl of the gradient is zero ($\nabla \times \nabla \Phi=0$), then if $$\nabla \times \vec B =\mu_0 \vec J$$ for the magnetic field $\vec B$, then we also have $$\nabla \times (\vec B+\nabla \Phi) =\mu_0 \vec J$$ for any (smooth) scalar field $\Phi$. This means that there is not a unique solution to the problem since $\vec B +\nabla \Phi$ is ...


1

This is simply because you are used to the regulardefinition of $\theta$ as being measured from the positive direction of the x-axis, as opposed to its definition here.


1

The good answer is indeed $\frac{4}{\pi}-\frac{1}{2}$. You can find a complete solution in the book of R. Lyons and Y. Peres "Probability on trees and networks", section 4.3, p. 124-127. This mainly uses Fourier analysis and the symmetry of the grid.


1

The short answer to your question is that, yes, you would have to apply a different transformation in your second example, but that, in that particular example, you would not always be able to determine with which of the two states the system had been in. The reason for the latter is that the two state vectors in the second example are not orthogonal, so no ...


1

I think he has a particular kind of coordinate system in mind: Choose a point $O$ as your origin, and then a basis $(e_0,e_1,e_2,e_3)$ for $\mathbb{R}^4$, where $e_0$ goes from $O$ to some point $P$ with $t(e_0)=t(\vec{OP})=1$, and where $(e_1,e_2,e_3)$ is an ON basis for the kernel of $t$. Then every event $Q$ (= point in $A^4$) has unique coordinates ...



Only top voted, non community-wiki answers of a minimum length are eligible