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Tutor profile: Eric H.

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Eric H.
High School Science Teacher with Background in Tutoring
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Questions

Subject: Basic Chemistry

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Question:

What is the electron configuration of a Strontium atom in the ground state?

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Eric H.
Answer:

$$1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^2$$ Or we can use the noble gas abbreviation where we go to the most recent noble gas and start our electron configuration there. $$[Kr]5s^2$$

Subject: Calculus

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Question:

Find the derivative of the following function: $$f(x) = \sin(x^2+x)$$

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Eric H.
Answer:

For this problem we will need to use the chain rule: $$\frac{d}{dx}(f(g(x)))=f^\prime(g(x)) \cdot g^\prime(x)$$. I know this looks messy but it will be our template when we try to find the answer. We use the chain rule when we have nested functions, or one function being placed inside the other. For this problem the expression $$x^2 + x$$ in inside the sine function. So we'll split it into two pieces. Our inner function is $$g(x) = x^2 +x $$. Our outer function is $$f(g) = \sin(g)$$ (Note that I've replaced the inside bit with the function $$g(x)$$). We now will need to find the derrivative of each piece. For the inner function we will use the power rule. $$g^\prime(x) = 2x+1$$. The outer function requires us to remember back on our memorized derivatives of the trig functions. $$f^\prime(g) = \cos(g)$$. Now going back to the definition of the chain rule we can plug stuff in: $$\frac{d}{dx}(f(g(x)))=\cos(g(x))\cdot(2x+1)$$. However if you notice we still have a $$g(x)$$ in there that we need to replace which gives our final derivative: $$\frac{d}{dx}(f(g(x)))= \cos(x^2+x)\cdot(2x+1)$$

Subject: Physics

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Question:

Consider a cannon on top of a 30 meter tall cliff that fires a projectile at 25 m/s, $$30 ^{\circ}$$ above the horizontal. How far from the base of the cliff does the cannon ball land? For this problem you may assume the air resistance to be negligible.

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Eric H.
Answer:

First we will split the problem into two seperate pieces, the horizontal piece and the vertical piece. We will get these values using trig to split the velocity vector into its separate components. $$v_x = v\cdot\cos\theta=(25 m/s)\cos(30^{\circ}) = 21.65 \text{ m/s}$$ $$v_y = {v}\cdot\sin\theta=(25 m/s)\sin(30^{\circ}) = 12.5 \text{ m/s}$$ Now, we need to figure out what forces our projectile is feeling. Since we are neglecting air resistance we know that the cannonball doesn't feel any force in the x direction. This means that the acceleration in the x direction is $$a_x=0 \text{ m/s}^2$$. The only force in the y direction is due to gravity, thus $$a_y = -g$$ in the downward direction (remember that $$g=9.81 \text{ m/s}^2$$). We're going to attack the vertical problem first because we are able to solve for the time that the projectile is in the air. We will use one of Newton's kinematic equations, $$x_f = x_0 + vt+ \frac{1}{2}at^2$$. Remember that the x in that equation just means position and has nothing to do with the xy coordinate system that we've created in this problem. Plugging the numbers that we know into that formula gives us the following expression. $$0 \text{ m} = 30 \text{ m} + (12.5 \text{ m/s})t - \frac{1}{2}gt^2$$ I'm going to use the quadratic formula to solve this quadratic: $$t = \frac{-12.5 \text{ m/s} \pm \sqrt{(-12.5 \text{ m/s})^2-4(\frac{-g}{2})(30 \text{ m})}}{2(\frac{-g}{2})}$$ Simplifying gives two solutions, $$t=4.06 \text{ s}$$ and $$t = -1.51 \text{ s}$$. Since a negative time doesn't make any sense in this context we will discard that solution. Now that we have a time value we can move on to the horizontal section of the problem. For this section we can use the distance formula for a body moving at a constant velocity: $$\Delta x = v\Delta t$$. Thus the cannonball will go $$\Delta x = 21.65 \text{ m/s} \cdot 4.06 \text{ s}$$ which gives us an answer of $$87.7 \text{ m}$$.

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