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Tutor profile: William B.

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William B.
Physicist having been a student assistant for several years
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Questions

Subject: Physics (Newtonian Mechanics)

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Question:

Inclined plane oscillator A heavy material point, of mass m, is forced to move on a line inclined by a angle $$\alpha$$ to the horizontal. There is no friction. The material point is retained by a spring of length at rest L and elastic constant k. a) Establish the balance of forces. b) Find the equation of motion. c) What is the period of the oscillations?

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William B.
Answer:

We place ourselves in the Cartesian reference frame Oxy. a) Balance of forces: Gravity $$\textbf{P} = m\textbf{g} = mg (sin \alpha \textbf{ex} - cos \alpha \textbf{ey})$$ The reaction of the support $$\textbf{R} = R\textbf{ey}$$ The tension of the spring $$\textbf{T} = -kx\textbf{ex}$$ b) There is no movement along Oy, which gives $$R = mg cos \alpha$$. Newton's second law according to Ox gives the equation of motion: $$m \ddot{x} = mg sin \alpha - kx$$ (1) We recognize the equation of a harmonic oscillator. By carrying out the change of variable $$u = x-\frac{mg}{k}sin \alpha$$, we fall back on the known writing: $$m \ddot{u} + ku = 0 $$ (2) c) The general solution of equation (1) is of the type $$x (t) = Acos (\omega t + \theta)$$. Which gives the condition $$\omega = \sqrt(\frac{k}{m})$$ With the relationship $$\omega = \frac{2\pi}{T}$$, we arrive at the solution $$T = 2\pi\sqrt(\frac{m}{k})$$

Subject: Statistics

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Question:

The following table provides the grades of students in a class during a math homework: \begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Grade} & 2 & 4 & 5 & 7 & 8 & 9& 10 & 11 & 12 & 14 & 15 & 18 \\\\ \hline \text{Effective} & 1 & 2 & 3 & 2 & 3 & 4 & 5 & 3 & 3 & 2 & 1 & 1 \\\\ \hline \end{array} 1) What is the percentage (to $$1\%$$ ) of students in this class who obtained a grade: a) between 8 and 12 (including values)? b. strictly less than 9 ? 2) Determine the range, median, quartiles of this series. 3) Determine the class average on this assignment. 4) In another class, there are 20 girls and 15 boys. At one control, the girls 'average was 11,8 and the boys' average was 10,2. What was the class average? 5) This control was common with the first class of 30 pupils, the average of the two classes was 10,7. What was the average in the first class?

Inactive
William B.
Answer:

1)a). 18 students on 30 have a grade between 8 and 12. This therefore represents $$\frac{18}{30} = 60 \%$$ of students. b). 11 students have a mark strictly lower than 9. This therefore represents $$\frac{11}{30} = 36,7 \%$$ of students. 2) The extent is 18-2=16. The median is the average of the 15ith and of the 16ith value, i.e $$\dfrac{9 + 10}{2} = 9,5.$$ $$\dfrac{30}{4} = 7,5$$. The first quartile is therefore the eighth value, i.e $$Q_1 = 7$$. $$\dfrac{30 \times 3}{4} = 22,5.$$ The third quartile is therefore the 23ith value, i.e $$Q_3 = 11$$. 3) The mean is $$\dfrac{2 \times 1 + 4 \times 2 + \ldots + 18 \times 1}{30} = 9,3$$ 4) The class average is $$\dfrac{20 \times 11,8 + 15 \times 10,2}{35} = \dfrac{389}{35} \approx 11,11 $$ 5) We call the $$x $$mean sought. So we have $$\dfrac{30x + 389}{30 + 35} = 10,7$$ So $$30x + 389 = 65 \times 10,7$$ Hence $$30x + 389 = 695,5$$ and $$30x = 306,5.$$ Therefore $$x = \dfrac{306,5}{30} \approx 10,22.$$

Subject: Physics

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Question:

A projectile is fired from the ground with an initial velocity $$v_0$$ at a firing angle $$\alpha$$ with respect to the horizontal. a) Establish the hourly equations along the x and y axes. b) Determine the angle $$\alpha^*$$ which maximizes the range $$x(\alpha) $$ shooting.

Inactive
William B.
Answer:

We choose the ground as reference frame, as reference system Cartesian coordinates in the vertical plane Oxy centered on the firing position, as system the projectile. a) Initial conditions: Position: $$(x_0, y_0) = (0, 0)$$ (1) Speed: $$(v_{0x}, v_{0y}) = (v_0 cos(\alpha), v_0 sin(\alpha))$$ (2) Law of motion: $$ \textbf{a = g}$$ (3) The equation of motion (3) projected along the x and y axes gives respectively $$\ddot{x} \equiv \frac{d^2x}{dt^2}=0$$, (4) $$\ddot{y} \equiv \frac{d^2x}{dt^2}= -g$$. (5) By integrating the equations of motion (4) and (5) with respect to time, we obtain the components of the speed along the x and y axes respectively, i.e. $$\dot{x} \equiv \frac{dx}{dt}= v_{0x}$$, (6) $$\dot{y} \equiv \frac{dx}{dt}= -gt + v_{0y}$$. (7) By integrating the components of the speed (6) and (7) with respect to time, we obtain the hourly equations along the x and y axes respectively, i.e. $$x(t)= v_{0x}t + x_0,$$ (8) $$y(t)=-\frac{1}{2}gt^2 + v_{0y}t + y_0.$$ (9) The initial conditions (1) and (2) imply that the hourly equations (8) and (9) according to the x and y axes are respectively equal, $$x(t)= v_{0}cos(\alpha)t,$$ (10) $$y(t)=-\frac{1}{2}gt^2 + v_{0}sin(\alpha) t$$ (11) b) The projectile hits the ground at y = 0. Equation (11) implies that this impact takes place at time $$t = \frac{2v_0}{g}sin \alpha$$ (12) By substituting this time in the hourly equation (10) along the x axis, we obtain the range of shooting, i.e. $$x(\alpha) = \frac{2v_{0}^{2}}{g}sin \alpha cos \alpha$$. (13) Using the trigonometric formula, $$sin (2\alpha) = 2 sin(\alpha) cos(\alpha)$$, (14) the range of the shot (13) becomes $$x(\alpha) = \frac{v_{0}^{2}}{g}sin 2\alpha$$ (15) To find the angle $$ \alpha^*$$ which maximizes the range of the shot, we cancel the derivative of equation (15) with respect to the angle $$\alpha$$, i.e. $$\frac{dx}{d\alpha}|_{\alpha^*} = \frac{v_0^2}{g}2cos(2\alpha^*)=0 \Rightarrow \alpha^* = \frac{\pi}{4}$$ (16)

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