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Tutor profile: Alex N.

Alex N.
Undergraduate Civil Engineering Student, Minoring in Electrical Engineering and Business Administration

Questions

Subject: Python Programming

TutorMe
Question:

What is the best way to remove the duplicates from the following list? groceries = ['milk', 'milk', 'carrots', 'pasta', 'pasta', 'coffee']

Alex N.
Answer:

One method would be to iterate over the list to find duplicates and remove them upon identification. However, the best method is to use the set type which doesn't allow for duplicates. To do so, the following function would be used. list(set(groceries))

Subject: Calculus

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Question:

Differentiate the following function: f(x) = (4x^3 + 3x^2)^5

Alex N.
Answer:

To solve this problem, it is crucial to use the "chain rule". Begin with the outside function, which is the exponent of 5. Therefore the inside function is (4x^3 + 3x^2). Having these defined, the derivative can be found as: f'(x) = 5 * (12x^2 + 6x) * (4x^3 + 3x^2)^4

Subject: Physics

TutorMe
Question:

An archer pulls his arrow back 0.85 meters on a bow whose stiffness is equal to 250 N/m. If the arrow weighs 75 grams, what is its velocity upon release?

Alex N.
Answer:

In order to simplify this problem, it helps to consider the bow as a spring. By approaching the problem in this manner and using Hooke's Law, the potential energy in the bow can be found through the equation PE = (1/2)*k*(x^2), where k is the spring constant and x is the distance the bow is stretched. Therefore, k = 250 N/m and x = 0.85 meters. Plugging these values in, the potential energy can be calculated to be: PE = (1/2)*(250)*(0.85^2) = 90.31 J. It is also known that PE = KE through energy conservation and the equation for kinetic energy is: KE = (1/2)*(mass)*(velocity^2). The mass is known to be 75 grams or 0.075 kg. Therefore the velocity of the arrow is equal to: 90.31 = (1/2)*(0.075)*(v^2) Through algebra, velocity is found to be equal to: v = 49.07 m/s.

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