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Tutor profile: Cristian M.

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Cristian M.
University Professor at Universidad Nacional de Colombia
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Questions

Subject: Mechanical Engineering

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Question:

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.

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Cristian M.
Answer:

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

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Question:

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.

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Cristian M.
Answer:

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

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Question:

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?

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Cristian M.
Answer:

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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