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Tutor profile: Tanner L.

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Tanner L.
Mechanical Engineering Student at NC State
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Questions

Subject: Trigonometry

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Question:

A right triangle has an angle angle of 23deg and a side length of 5.2m adjacent to it. Find the hypotenuse of the triangle.

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Tanner L.
Answer:

This problem requires the use of the inverse cosine trigonometric function. We know that the cosine function can be written as such: $$ \cos{\theta} = \frac{adjacent}{hypotenuse} $$ Plugging in what we know, we get the following. For the purposes of this problem, we'll refer to the hypotenuse length as $$ H $$: $$ \cos(23\deg) = \frac{5m}{H} $$ Rearranging the above equation to isolate $$ H $$, we get the following: $$ H = \frac{5m}{\cos(23\deg)} $$ Finally solving the above equations gives: $$ H = 5.43m $$

Subject: Mechanical Engineering

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Question:

Using a 2D x and y plane as a frame of reference, there are 3 forces that all act at the origin (0,0). One force of 50N acts at $$60^o$$ from the positive x-axis and another force of 30N acts at $$120^o$$ from the positive x-axis. Find the magnitude and direction of the third in order to maintain equilibrium.

Inactive
Tanner L.
Answer:

The best way to approach this problem is to break the given forces into x and y vectors, then summing the forces in the x and y direction to find the required vectors to maintain equilibrium. First, lets break the given forces into vectors: Force 1 x-direction: $$(50N)\cos{60} = 25N $$ y-direction: $$ (50N)\sin{60} = 43.3N $$ Force 2 x-direction: $$(30N)\cos{120} = -15N $$ y-direction: $$ (30N)\sin{120} = 25.9N $$ Now summing the forces in the x and y direction gives: $$ \sum{Fx +25N -15N} $$ $$ Fx = 15N -25N $$ $$ Fx = -10N $$ $$ \sum{Fy +43.3N + 25.9N} $$ $$ Fy = -43.3N - 25.9N $$ $$ -69.2N $$ Now we just need to convert these x and y components into a magnitude with a direction: $$ \vert F \vert = \sqrt{(-10N^2) + (-69.2N^2)}$$ $$ \vert F \vert $$ = 69.92N $$ \arctan({\frac{-69.2N}{-10N})} = 81.77\deg + 180\deg = 261.78\deg$$ *The reason for adding 180deg to our final angle is because our x and y components of our force vector show the vector acting in the third quadrant of the plane, therefore 81.77deg would not make sense.

Subject: Calculus

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Question:

Find the following indefinite integral: $$ \int\mathrm{6x }^{5} - 18x^2 + 7\mathrm{d}x $$

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Tanner L.
Answer:

To make solving this integral easier, we can look at this as three separate integrals by breaking up each of the individual terms like so: $$ \int\mathrm{6x }^{5}\mathrm{d}x - \int_0^\infty \mathrm18x^2\mathrm{d}x + \int_0^\infty7\mathrm{d}x$$ Analyzing the first integral, we simply raise the power by one and divide by the new value like so: $$ \int\mathrm{6x }^{5}\mathrm{d}x = \frac{6x^6}{6} = x^6 $$ Analyzing the second integral using the same method: $$ -\int\mathrm{18x }^{2}\mathrm{d}x = \frac{-18x^3}{3} = -6x^3 $$ The third term is an integer value with no variable, so we can simply add an $$x$$ to this term like so: $$ \int\mathrm{7} \mathrm{d}x = 7x $$ Now, because this is an indefinite integral, we must also add a constant variable to the end of our solution to represent the possibility of a constant in our general solution. With this in mind, the final solution becomes: $$ x^6 - 6x^3 + 7x + c $$

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