# Tutor profile: Thomas R.

## Questions

### Subject: Physics (Newtonian Mechanics)

A football kicker is starting a game. Immediately after the kick, the ball has an initial speed of 18 m/s at 30 degrees above horizontal. The initial angular speed of the ball is 600 rpm. The ball may be approximated as a thin rod 0.1m long with a mass of 0.4kg. Air resistance may be neglected and the ball will not lose any rotational or horizontal speed. 1) Write the kinematic equations that describe the ball's position. 2) How far will the ball travel? 3) How high will the ball travel? 4) What is the hang time? 5) What is the total energy of the ball at the top?

1) 2) 3) We can assume from the problem that the ball will start and end at the same height. Additionally, we are told that air resistance can be neglected and that the ball will not lose any rotational or horizontal speed. Therefore, we can use the range equation in order to calculate horizontal range: Range = $$\frac{v_{i}^2sin2\theta}{g}$$ In this case, $$sin2\theta = sin(2*30) = sin(60) = \frac{\sqrt{3}}{2}$$ $$g$$ can be approximated as 9.81 m/s$$^2$$ (note that this is a positive value as $$g$$ is the magnitude of acceleration due to gravity. Substituting the rest of our values: Range = $$\frac{(18 m/s)^2 * \sqrt{3}}{(9.81 m/s^2) * 2}$$ $$\textbf{Range = 28.7 m}$$ (Note that this could have been done directly from the kinematic equations in part 1), but using the range equation saves some time and effort.) 4)

### Subject: Biology

1) Track the steps of calcium influx into the cell from the initial ligand binding to activation of transcription factors. 2) Consider a hypothetical patient that has a mutation in the IP$$_{3}$$-gated Ca$$^{2+}$$ channels on the endoplasmic reticulum that does not allow the channels to open. Explain how this mutation may potentially impact skeletal muscle contractions. Be specific regarding the role of calcium in the cellular mechanism of muscle contraction.

1) A variety of ligands may activate the calcium release pathway, but all of them start by activating a G-protein couple receptor. This GPCR activates $$\alpha$$ and $$\beta/\gamma$$ subunits through phorphorylation, which in turn activate phospholipase C-$$\beta$$. Phospholipase C-$$\beta$$ catalyzes the conversion of phosphatidylinositol 4,5-bisphosphate (PIP2) to inositol 1,4,5-triphosphate (IP$$_{3}$$). IP$$_{3}$$ then opens ligand-gated channels on the endoplasmic reticulum that allow stored calcium to be released into the cytosol. This initial calcium release changes the conformation of the STIM protein in the ER. After translocation to the membrane, the STIM protein opens the ORAI-1 calcium channel on the plasma membrane, allowing calcium to enter the cell. Calcium then has a variety of intracellular targets with a wide variety of functions. 2) If the IP$$_{3}$$ channels are unable to open (allowing for the release of stored calcium), the STIM protein will never change conformation and open the ORAI-1 transmembrane calcium channel. Although low levels of calcium may remain in the cell, the transcellular calcium gradient cannot be utilized to increase intracellular calcium concentration. In a healthy patient, the influx of calcium would allow for the binding of troponin, a complex which can remove tropomyosin from the myosin binding sites on actin filaments, allowing for myosin walking. However, a lack of calcium means that troponin will not be activated and tropomyosin will remain on the actin filament binding sites. Therefore, this patient would likely have difficulty in skeletal muscle actuation, as opposed to something like prolonged muscle contraction (tetanus).

### Subject: Algebra

Consider the two lines: Line 1: $$y = x - 3$$ Line 2: $$y - 3 = 3(x + 2)$$ 1) What is the $$x$$-intercepts, $$y$$-intercept, and slope of Line 1? 2) What is the $$x$$-intercept, $$y$$-intercept, and slope of Line 1? 3) What is the point of intersection of Line 1 and Line 2? Give your answer as a standard coordinate pair ($$x, y$$).

The lines given in the question are given in two different forms The Line 1 is given in slope intercept form $$y = mx + b$$ and the Line 2 is given in point slope form $$y_{2}- y_{1} = m(x_{2} - x_{1})$$. 1) By simply examining the equation for Line 1 $$(y = x - 3)$$, we can easily determine the slope and $$y$$-intercept: $$\textbf{the slope (m) is 1}$$ (since x does not have a written coefficient, you can infer that the coefficient is 1) and that $$\textbf{the y-intercept (b) occurs at y = -3}$$. The $$x$$ intercept occurs where the line crosses the $$x$$-axis, so that can be calculated by finding the $$x$$ value when $$y$$ is equal to zero: $$0 = x - 3$$ $$x = 3$$ Therefore, $$\textbf{the x-intercept occurs at x = 3}$$. Note that this is $$+3$$ and not $$-3$$. 2) The slope of Line 2 may also be determined just by examining the equation of the line $$y - 3 = 3(x + 2)$$. The coefficient of the $$(x-2)$$ term is, by definition, the slope, so we can see from the equation of Line 2 that $$\textbf{the slope is equal to 3}$$. The $$x$$ and $$y$$ intercepts may not be immediately apparent from the equation, but we can use the same logic as part 1). To find the $$x$$-intercept, you need to find the point at which $$y = 0$$. Therefore, we can set $$y = 0$$ in the Line 2 equation: $$y - 3 = 3(x + 2)$$ $$0 - 3 = 3(x+2)$$ $$\frac{-3 = 3(x+2)}{3}: -1 = x+2$$ $$x = -3$$ Therefore, $$\textbf{our x-intercept occurs at x = 3}$$. The same process can be used to find the $$y$$-intercept by setting $$x = 0$$: $$y - 3 = 3(x + 2)$$ $$y - 3 = 3(0 + 2)$$ $$y - 3 = 3(2) = 6$$ $$y = 9$$ Therefore, $$\textbf{our y-intercept occurs at y=9}$$ 3) The intersection of two lines occurs when they have the same values for $$x$$ and $$y$$. Since Line 1 is already in the form of $$y=mx+b$$, we can manipulate Line 2 to be in this form and then set them equal to each other: $$y - 3 = 3(x + 2)$$ $$y = 3(x + 2) + 3$$ Distribute the $$3$$ over the $$x + 2$$ term: $$y = 3x + 6 + 3$$ $$y = 3x + 9$$ (Line 2 is now in slope-intercept form). We can now find the $$x$$-value of our intersection point by setting the two equations equal to each other: Line 1: $$y = x - 3$$ Line 2: $$y = 3x+9$$ $$x - 3 = 3x+9$$ Now, get all of the terms with $$x$$ onto one side: $$[x - 3 = 3x + 9] -x$$ $$-3 = 3x-x+9$$ $$-3 =2x+9$$ Get all of the constant terms onto the other side: $$[-3 =2x+9] - 9$$ $$-12 = 2x$$ $$\frac{-12}{2} = \frac{2x}{3}$$ $$-6 = x$$ We can now use this value to find the $$y$$-value of our intersection point by plugging in $$x = -6$$ to the equation for either line. We will use Line 1 here: $$y = x - 3$$ $$y = -6 -3$$ $$y=-9$$ Therefore, $$\textbf{the point of intersection occurs at (x, y) = (-6, -9)}$$

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