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Hans P.
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Physics (Newtonian Mechanics)
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Question:

A block is sliding across a wood beam with a coefficient of friction of .2. The block weight 5kg and was accelerated to the right until it reached the velocity of 5 meters/second. How long does it take for the block to slow to a complete stop?

Hans P.
Answer:

First we label what we know. We know the block weights 5kg so the force of gravity is 5kg * 10 m/s/s or 50 Newtons. We also know that the block is on a level surface so our Normal force is also 50 Newtons. The force of friction (Ff) is equal to the coefficient of friction (mu) * normal force(N) so .2 * 50 = 10 Newtons the sum of all forces = mass * acceleration our force of friction is acting on the opposite direction of our movement so it is negative relative to us. -Ff = ma -(mu*N) = ma a = (-mu*N)/m a = (-mu*m*g)/m a = (-mu * g) a = -.2 * 10m/s/s a = -2m/s/s Now we can use our movement equations to find out when the block is stopped Vf = Vo + at 0 = 5 - 2*t 2t = 5 t = 2.5 seconds.

Java Programming
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Question:

Write a program that prints the numbers from 1 to 100. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz"

Hans P.
Answer:

public class FizzBuzz { //the main method that is called public static void main(String[] args) { //For loop that increments i from 1 to 100 for (int i = 1; i <= 100; i++) { //Multiple of 3 and 5 because the mod operator returns 0 if (i%3 == 0 && i%5 == 0) System.out.println("FizzBuzz"); //If not multiple of 3 and 5, then is it a multiple of three else if (i%3 == 0) System.out.println("Fizz"); //Well since it's not a multiple of 3 and 5, and it's not a multiple of 3, is it a multiple of 5? else if (i%5 == 0) System.out.println("Buzz"); //Well, it's not a multiple of 3 or 5, time to just print the number. else System.out.println(i); } }

Calculus
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Question:

A 20 foot ladder is resting against the wall. The bottom is initially 15 feet away from the wall and is being pushed towards the wall at a rate of 1/2 ft/sec. How fast is the top of the ladder moving up the wall 12 seconds after we start pushing?

Hans P.
Answer:

This is a related rates question. We can use Pythagorean theorem and it's derivative in order to calculate how fast the top of the ladder is moving up the wall at t=6 (dA/dt) To start we know the ladder is 20 feet long and it will always be 20 feet so long so it is a constant. Hence we can substitute 20 in for c in a^2 + b^2 = c^2. Therefore a^2+b^2 = 20^2 or 400 At t = 0, we know the ladder is 15 feet away from the wall and we know that it is moving in .5 feet every second so after 6 seconds it will have moved in (6 second * 1/2 feet/second) so three feet. Using this we determine that our length is 15 - 3 = 12 ; using this we can find out how high up the ladder is on the wall using the Pythagorean Theorem. 20^2 - 12^2 = a^2 256=a^2, therefore a = 16 Now we take the derivative of the original function f = a^2 + b^2 = 400 f' = 2a(da/dt) + 2b(db/dt) = 0 We know the ladder is being pushed towards the wall at .5 feet/sec, so db/dt is -.5 feet/sec and we can fill in b = 15 and a = 13.229 in from our previous calculations. 2(16)(da/dt) + 2(12)(-.5) = 0 32(da/dt) - 12= 0 32(da/dt) = 12 da/dt = ..375 ft/sec Therefore, we know that the ladder is moving up the wall at t=12 at .375 feet per second. *****NOTE******** This is a question best done with visuals and drawing pictures to show the change.

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