# Tutor profile: Cristian M.

## Questions

### Subject: Mechanical Engineering

Let us consider a Dynamic System of equations of a mass and damper system with two masses defined by the equations $$\displaystyle 15\ddot{x_1}+7\dot{x_1}-4\ddot{x_2}+30x_1-15x_2=0$$ $$ \displaystyle 6\ddot{x_2}-15x_1+15x_2-4\ddot{x_1}+4\ddot{x_2} = 9f(t)$$ Define the state variables models in matrix mode.

We defined the state equations and the output equations in general form as, For the case of state equation we have that, $$\dot{x} = \mathbf{A}\cdot x + \mathbf{B}\cdot u$$ Here, x = State Vector u = Input Vector A = State Matrix B = Input Matrix For the output equations we have that, $$y = \mathbf{C} \cdot x + \mathbf{D}\cdot u$$ Here, y = Output Vector C = Output Matrix D = Direct Tranmission Matrix The state variable models are $$\displaystyle \dot{x_1}=x_3\\ \displaystyle \dot{x_2}=x_4\\ \displaystyle \dot{x_3}=\frac{1}{15}(-7x_3+4x_4-30x_1+15x_2)\\ \displaystyle \dot{x_4}= \frac{1}{6}(4x_3-4x_4+15x_1--15x_2+9f(t))$$ Using the Matrix form, $$\begin{bmatrix} \dot{x_1}\\ \dot{x_2}\\ \dot{x_3}\\ \dot{x_4} \end{bmatrix} = \begin{bmatrix} 0& 0& 1& 0\\ 0& 0& 0& 1\\ -2& 1& -\frac{7}{15}& \frac{4}{15}\\ \frac{5}{2}& -\frac{5}{2}& \frac{2}{3}& -\frac{2}{3}\\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}+\begin{bmatrix} 0\\ 0\\ 0\\ \frac{9}{6} \end{bmatrix}f(t) $$ Finally we have that output vectors as, $$\begin{bmatrix} y_1\\ y_2 \end{bmatrix}=\begin{bmatrix} 1&0&0&0\\ 0&1&0&0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}+0$$

### Subject: Astronomy

Hubble's law $$v = Hd$$, represents a linear approximation of the displacement of galaxies. In this equation, $$v$$ represents the velocity of the galaxy, $$H$$ the Hubble constant and d the distance between them. If we consider a certain distance equivalent to $$d$$ between two galaxies which move at a speed given by Hubble's law, and provided that the galaxies do not escape to infinity: Determine the critical density.

According to the previous definition we have define that, $$v = H_0 D \Rightarrow \text{Eq. 1}$$ Here, $$\mathbf{v} = \text{Recessional velocity (km/s)}\\ \mathbf{H_0} = \text{Hubble's constant (Hubble parameter)}\\ \mathbf{d} = \text{Proper distance (Megaparsecs Mpc)}$$ Similarly the critical density is defined, considering the Hubble constant and the universal constant of gravitation, as $$\displaystyle \rho_c = \frac{3H_0^2}{8\pi G} \Rightarrow \text{Eq. 2 }$$ Here, Rearranging Equation 1 in terms of Hubble's constant $$ \displaystyle H_0 = \frac{v}{D} $$ Then replacing at equation 2, $$\displaystyle \rho_c = \frac{3\left(\frac{v}{D}\right)^2}{8\pi G}\\ \boxed{\therefore \mathbf{\rho_c} = \frac{3v^2 }{8\pi GD^2}}$$

### Subject: Physics

Suppose that at a certain moment the sun collapses and becomes a sphere of radius R. The energy necessary to be able to remove a fraction $$m$$ of its mass is equivalent to $$mc ^ 2$$. The new radius $$R$$ of the sun is theoretically considered the gravitational radius, what would that new radius be?

The ideal concept to solve this problem is the conservation of energy in which we will equal the work done to remove the mass to the potential energy, $$\displaystyle W = \frac{GM_sm}{R}$$ Here, $$\mathbf{G} = \text{Universal gravitational constant}\\ \mathbf{M_s} = \text{Mass of the sun}\\ \mathbf{m} = \text{Mass removed}\\ \mathbf{R} = \text{Radius of the Sphere}$$ The work is equal to the energy necessary to remove the mass, then $$W = mc^2$$ Here, $$\mathbf{m} = \text{Mass removed}\\ \mathbf{c} = \text{Speed of the light}$$ If we know equating, $$ \displaystyle \frac{GM_sm}{R}=mc^2 \\ \displaystyle R = \frac{GM_s}{c^2}$$ Replacing, $$\displaystyle R = \frac{(6.67*10^{-11}m^3/kg \cdot s^2)(1.99*10^{30}kg)}{(3.00*10^8m/s)^2}\\ \displaystyle R = 1.47*10^3 m \\ \displaystyle \boxed{\therefore \mathbf{R} = 1.47km} $$ Therefore the radius of the sphere will be equal to 1.47km

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