Enable contrast version

# Tutor profile: Alex N.

Inactive
Alex N.
Tutor Satisfaction Guarantee

## 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']

Inactive
Alex N.

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

TutorMe
Question:

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

Inactive
Alex N.

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?

Inactive
Alex N.

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.

## Contact tutor

Send a message explaining your
needs and Alex will reply soon.
Contact Alex

Start Lesson

## FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage