# Tutor profile: Andrew L.

## Questions

### Subject: Geometry

A straight line passes through the points $$\left(-1,-4\right)$$ and $$\left(5,2\right)$$. Find the equation of the line in the form $$y=mx+c$$.

The starting point with this sort of question is always to find the gradient, $$m$$. We are working with a straight line, so the gradient is simply the change in $$x$$ divided by the change in $$y$$: $$m=\frac{\Delta{x}}{\Delta{y}}=\frac{x_2 - x_1}{y_2 - y_1}$$ Substituting in the known coordinates gives: $$m=\frac{5-(-1)}{2-(-4)} = \frac{6}{6} = 1$$ Now we have the gradient we have a modified form for the line equation: $$y = 1 x + c = x+c$$ All that is left is to find c, by inserting either coordinate into the equation. Using $$\left(-1,-4\right)$$ gives: $$-4 = -1 + c$$ $$c = -3$$ Leaving us with the final equation: $$y = x -3$$

### Subject: Basic Chemistry

Magnesium can react with the oxygen in the air to create magnesium oxide. Write a fully balanced equation for this reaction.

The first step to create the reaction equation is to correctly identify the individual components. Magnesium is in elemental form, so can simply be written as Mg. Oxygen in the air is diatomic, so is given the symbol O$$_2$$. Magnesium oxide is the oxide formed when magnesium and oxygen combine. Magnesium is in group 2 of the periodic table, so will form the Mg$$^{2+}$$ cation, whereas oxygen is in group 16, so will form the O$$^{2-}$$ anion. To balance these charges, one magnesium ion combines with one oxygen ion to give the compound MgO. We can now write out the reaction equation as: $$\textrm{Mg}+\textrm{O}_2 \rightarrow \textrm{MgO}$$ However, this is not balanced, as we have more oxygen atoms on the left than the right. To solve this, we need to add an extra MgO molecule: $$\textrm{Mg}+\textrm{O}_2 \rightarrow 2\textrm{MgO}$$ But now the magnesium atoms don't balance! So, adding another magnesium to the left gives: $$2\textrm{Mg}+\textrm{O}_2 \rightarrow 2\textrm{MgO}$$ and the equation is fully balanced

### Subject: Physics

The rise and fall of the tides can be approximately modeled using simple harmonic motion (SHM). If the water is 4.0m deep at low tide, and 7.0m deep at high tide 6 hours later, calculate the period and amplitude of the SHM.

Period: The period is the time taken for one complete cycle of the SHM to complete. In this case, we can take the time between consecutive low tides to give us the period. We are told that it takes 6 hours to go from low tide to high tide. As we are dealing with SHM, the time taken to go from high tide back to low tide again must be the same, so the period is: $$ T = 6 \textrm{ hours} + 6\textrm{ hours} = 12 \textrm{ hours}$$ Amplitude: The amplitude is the greatest distance moved away from the equilibrium position (i.e. the middle of the path). For this question, the equilibrium position is at 5.5m (halfway between the high and low tide, which we are given as 7.0m and 4.0m respectively). The amplitude is then the difference between the equilibrium position and the depth and either the low or high tide: $$ A = 7.0\textrm{ m} - 5.5\textrm{ m} = 1.5\textrm{ m}$$

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