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Ryan K.
Computer Engineering Student Cal Poly San Luis Obispo
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Physics (Newtonian Mechanics)
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Question:

An object with mass 10 kg is moving 35 m/s and explodes into two smaller pieces of mass 7kg and 3kg. Ignoring energy lost to the explosion, calculate the velocity of the 3kg piece if the 7kg piece is moving 20m/s?

Ryan K.
Answer:

Using conservation of energy, we can calculate the total kinetic energy before the explosion with KE = (1/2)mv^2 = 6,125 Joules. Then knowing that the total kinetic energy after the explosion will be equal to the total kinetic energy before the explosion, we add the KE of each piece and set it equal to 6125. Then solve for v. 6125 = (1/2)(7)(20^2) + (1/2)(3)(v^2) v = 56.124 m/s

Java Programming
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Question:

Compare the efficiency of using a Linked List versus an Array.

Ryan K.
Answer:

The Linked list has the following advantages: The size of arrays are fixed. Thus we must know the max limit of elements before we create the array or else we will need to reallocate memory. In addition, the memory is reserved and often times goes unused, unless the array actually gets filled. Inserting an element into an array is costly. If an element is to be added first, all elements must be shifted down. In linked list, only simple pointer reassignment is needed. The Linked List has the following disadvantages: You can not randomly access elements, you must access elements starting from the first. In an array, random access is a very cheap operation. Arrays have better memory localization. When you allocate the size of your array, a chunk of memory is reserved for it all right next to each other. Linked list have pointers to all different places in memory.

Calculus
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Question:

Two ants are at a common point at time t=0, the first ant starts crawling along a straight line at the rate of 4 ft/min. Two minutes later, the second ant starts crawling in a direction perpendicular to that of the first, at a rate of 5 ft/min. How fast is the distance between them changing when the first insect has traveled 12 feet?

Ryan K.
Answer:

The key to this question is the Pythagorean theorem. With this we can relate the distance traveled by each bug to the distance between them. Taking a derivative with respect for time and then solving for dr/dt (r being the hypotenuse) and using the initial conditions provided in the problem we can work the problem out and determine the distance between them is changing at a rate of 5.6 ft/min.

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