# Tutor profile: Martin B.

## Questions

### Subject: Physics (Newtonian Mechanics)

Sir Isaac Newton dropped an apple from the top of a building. If it took the apple 2.5 seconds to reach the floor, how high was the building?

.Since the apple is acted upon only by the force of gravity while it's in the air, we can use our free-fall formulas to approach this problem. In particular, if we consider the formula for the height of an object experiencing free-fall motion: $(h = h_{0} + v_{0}t - \frac{1}{2}gt^{2} $) we can set $$ t = 2.5s $$ and $$ h = 0 $$ since we're told that the apple hits the floor $$(h = 0)$$ at 2.5 seconds after it's dropped. Also, since the problem says the apple is dropped its initial velocity $$v_{0}$$ is $$ 0$$. If we replace these values in the formula we're left with an equation for $$h_{0}$$, which is the height of the building: $(0 = h_{0} + (0)t - \frac{1}{2}(9.8 \frac{m}{s^2})(2.5s)^{2} $) From this equation we can isolate $$h_{0}$$ as follows: $(h_{0} = \frac{1}{2}(9.8 \frac{m}{s^2})(2.5s)^{2} $) and therefore: $(h_{0} = 30.63m$) The building is $$30.63m$$ tall.

### Subject: Calculus

.Find the equations of all lines that are tangent to the curve given by the equation $$ y = x^2 + 2 $$ and pass through the origin (0,0).

.The equation of the tangent line to a curve given by the function $$ y = f(x) $$ is given by: $(y = f(x_0) + f'(x_0)(x - x_0) $) In this case $$ f(x_0) $$ can easily be obtained by replacing $$ x $$ with $$ x_0 $$ in the equation of the curve: $( f(x_0) = x_0^2 + 2 $) and in order to find $$ f'(x_0) $$ we just need to take the derivative of $$ y = x^2 + 2 $$ and evaluate it at $$ x_0 $$: $(f'(x) = 2x $) $(f'(x_0) = 2x_0 $) Plugging both of these values in the equation of the tangent line given above we get: $(y = f(x_0) + f'(x_0)(x - x_0) $) $(y = x_0^2 + 2 + 2x_0 (x - x_0) $) The expression above represents the equation of a line that is tangent to the given curve at a generic point $$ x_0 $$, but the problem also asks that we select only the lines that pass through the origin. To enforce this we just need to ensure that the equation of the line meets the requirement that y equals 0 when x is set to 0. We can enforce this by setting x equal to 0 and y equal to 0, which yields: $( 0 = x_0^2 + 2 + 2x_0 (0 - x_0) $) Note that this is actually an equation for $$ x_0 $$. In other words, solving this equation will give us the actual value(s) of $$ x_0$$ for which the above requirements are met. This equation is easy to solve: $( 0 = x_0^2 + 2 - 2x_0^2 $) $( 0 = - x_0^2 + 2 $) $( x_0^2 = 2 $) $(x_0^2 \pm \sqrt{2} $) This means there are two values of $$ x_0 $$ that meet the requirements: $( x_1 = \sqrt{2} $) $( x_2 = - \sqrt{2} $) and therefore two lines associated with them. Since the problem asks for the equations of the lines (not just the values of $$ x_0$$) we just need to plug these values in the formula for the tangent line. For $$ x_1 $$ we get: $( y = f(x_1) + f'(x_1)(x - x_1) $) $( y = (\sqrt{2})^2 + 2 + 2\sqrt{2}(x - \sqrt{2}) $) $( y = 4 + 2\sqrt{2}x - 2\sqrt{2}\sqrt{2} $) $( y = 2\sqrt{2}x $) For $$ x_2 $$, we get: $( y = f(x_2) + f'(x_2)(x - x_2) $) $( y = (-\sqrt{2})^2 + 2 + 2(-\sqrt{2})(x + \sqrt{2}) $) $( y = 4 - 2\sqrt{2}x - 2\sqrt{2}\sqrt{2} $) $( y = -2\sqrt{2}x $) To summarize, there are two lines that meet the requirements, and their equations are given by: $( y = \pm 2\sqrt{2}x $)

### Subject: Physics

What is the wavelength of a photon whose energy is 3.1 eV? Does it correspond to visible light, and if so, which color is associated with it?

There’s a very straightforward formula that connects the energy of a photon with its wavelength: $(E = \frac{hc}{\lambda} $) where $$ h $$ is Planck’s constant ( $$ h = 6.626 \times 10^{-34} Js $$), $$ c $$ is the speed of light in vacuum ( $$ c = 3 \times 10^8 \frac{m}{s} $$), and $$ \lambda $$ is the wavelength. We can rearrange this formula to get an expression to compute the wavelength in terms of the energy: $( \lambda = \frac{hc}{E} $) Since Planck’s constant is expressed in $$ Joules \cdot second $$ and we have the energy of the photon expressed in $$ eV $$ we first need to convert the energy of the photon from $$ eV $$ to $$ Joules $$. Since we know that $$ 1 eV = 1.602 \times 10^{-19} J $$, we can do the conversion as follows: $( E = 3.1 eV = 3.1 eV \times \frac{1.602 \times 10^{-19} J}{1 eV} $) $(E = 4.97 \times 10^{-19} J $) Now we can go ahead and plug all the values in: $(\lambda = \frac{hc}{E} $) $( \lambda = \frac{6.626 \times 10^{-34} Js \times 3 \times 10^8 \frac{m}{s}}{4.97 \times 10^{-19} J } $) And simplifying this expression we get: $( \lambda = 4.00 \times 10^{-7} m = 400 nm $) which means the wavelength of the photon is 400 nanometers. By looking at a spectrum chart we can see that this photon’s wavelength falls within the range corresponding to visible light (380 – 750 nm), and in particular within the range corresponding to violet light (380 – 440 nm).

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