Tag: problem

Fascinating Frequencies 2

This challenge involves orbits.

A geostationary orbit is one where the satellite stays in the same position relative to the ground. This works because the earth spins; the velocity that keeps it over the same spot increases while the velocity needed to orbit decreases. At some point, they intersect. All geostationary orbits must be at the equator, and have the same altitude althroughout. If you wanted to get a satellite into geostationary orbit around a planet with mass of 5,000 yottagrams that completes 1 revolution every 20 hours, how fast would your orbital speed be at apoapsis? The apoapsis is the highest point in an orbit.

solution

Fascinating Frequencies 1

This challenge is the first of a new science challenge series, Fascinating Frequencies. You will need to use physics to solve this challenge.

Some radioactive particles are decaying. They are decaying in different ways, but each type of decay releases the same amount of energy. If the particle undergoes beta decay, it will release an electron and positron, both traveling at half the speed of light in a vacuum. If it undergoes gamma decay, it will release a photon. The challenge is to find the frequency of this photon.

solution

Peculiar Products

Trigonometry, geometry, ad algebra are required to solve todays challenge.

Pi notation is to Sigma notation as plus is to minus. Sigma notation is the sum of a bunch of stuff, Pi is the product. Pi is also ≈3.1416, but we will note that as π. In Strange Sums, we evaluated an expression with 2 Sigmas. Here will will find the product of the outputs of an equation. Specifically, the equation for the area of a regular polygon. Better go find that one out right now if you don’t know it. We will express this function as A(n, s), where n is the number of sides and s the length of them. Now solve the expression below.

\prod_{k=1}^\infin\Bigg(A\bigg(k,\,\frac1{tan (90\degree)-\frac{360\degree}k}\bigg)\Bigg)
Solution

Exquisite Equation

Advanced algebra is required to solve this challenge. You should read Perplexing Protons before beginning this challenge.

We left an unsolved equation after solving Perplexing Protons. In this challenge, we will solve it. Good luck!

x=5\,000\,000+E^+\frac{2-e^ {\frac{-1}{N^+}}-e^\frac{-3}{N^+}}{N^+} +E_{t=3.25}-3E^+-1\\[8pt] \frac{E^+-E^+e^\frac{-3}{N^+}} {N^+e^\frac1{N^+}}=1\,000\,000
Solution

Confounding Crystals

Chemistry knowledge is needed to attempt this challenge.

Crystals are very interesting and beautiful. There are seven types of crystal shapes. Plagioclase feldspar makes triclinic crystals. You have one that’s a prism whose faces have areas 30 cm2, 35 cm2, and 42 cm2. The sides have lengths 3cm, 11cm, and 14cm. The challenge is to use what you know about triclinic crystals to determine what the angles of the crystal must be. This challenge may seem to be about math, but really it’s just a little bit of trig. The important part is that you know the defining features of triclinic crystals.

Solution

Bouncy Ball

Physics simulations are highly reccomended to solve this science challenge.

You release a bouncy ball from 1 metre above the Earth. Neglect decrease from surface gravity. It will fall and bounce, then fall again, and so on.With each bounce, the ball’s velocity after the bounce is the negative root of it’s speed before it. For example, if it’s velocity was 4 m/s down, it’s velocity would become -2 m/s down, or 2 m/s up. Clearly, this is not a normal bouncy ball. In fact, it breaks several laws of physics. But still, calculate the velocity of the ball after 100 jumps.

Strange Sums

Algebra and geometry are needed to solve this challenge.

Sigma notation is the topic of today’s challenge. At the bottom, we set a variable, usually ‘K’. K typically starts at 0 or 1. We would write ‘k=0’ or ‘k=1’ below the Sigma. Sigma notation is often used to denote the partial or full sum of an infinite series. At the top, we say when the series will end, often infinity. The value of the expression is the sum of the stuff to the right of the sigma for every value of K. The challenge for today is to find the sum of the expression. Look out for the sequel to this challenge, Peculiar Products, in which we will discuss Pi notation.

\sum_{k=2}^{\infin}\Bigg(\sum_{l=0}^{\infin} \Big(\frac1{k^l}\Big)\Bigg)
Solution

Perplexing Protons

You will need to use chemistry principles to solve this challenge.

You have some protons in a particle collider, but you don’t know how many. Your computers said there were 5 million at 12 noon based on their magnetic field. The problem is that you just found out that some electrons were leaking into the collider. You discovered the leak and patched it at 2pm. You know the electrons were leaking since 11 o’clock, and that they leaked at a constant rate. There are now 1 million electrons at 2:15, and they disrupted the magnetic field from which you counted the protons. However, some electrons collided with protons to make neutrons. You need to know how many protons there are for your experiment at 2:30. You don’t have time to count all the neutrons or recount the protons. You must figure it out based on the amount of energy released, which was 1 kilojoule. The challenge is to use the information above to determine the number of protons that will be in the collider at 2:30 if the rate of protons colliding with electrons is linearly proportional to the number of electrons.

solution

Complex Coins

You will need to use algebra and combinatorics to solve this challenge.

A normal coin has 2 flat sides and one thin round edge. When tossed, the coin always lands on a flat side. A brand new pencil has 2 small flat sides and 1 long round side. When tossed, the pencil always lands on the round side. Somewhere in the middle is the 3 sided coin, where all 3 sides are equally likely to be rolled. You happen to have 5 such 3 sided coins*. You call the 3 sides heads, tails, and round, or H, T, and R for short. You and your friend John a game with these coins. For every H, you get a point. For every T, John gets a point. But for every R, you both get a point. First to 5 points wins. You start playing, and roll 2 tails and a round. What is the probability that you win?

*Despite various lengthy efforts, nobody has been able to produce a perfect 3 sided coin as described in this challenge. So far. Some have come close, see “How thick is a three-sided coin?” from Standupmaths on Youtube, available below.

Solution

Vile Voting

Algebra and combinatorics are needed to solve this challenge.

Brutus and Craw are voting over their preference of colours. Their satisfaction is measured by how close their preferences are to the ones agreed on; they do not care about each other. The vote is determined by each player prioritizing the colours of the rainbow. Brutus and Craw do not know about indigo. Their first pick gets 5 points, their second 4, and so on such that their last gets 0.The colour with the most is first, the second most, second, and so on. Brutus and craw receive “points” of satisfaction, based on how close the vote is to their personal preferences. For each preference the number of points starts at one less than the previous, with 5 for 1st. It goes down by 2 if it’s below it’s position, and then decreases by one until it stops at 0. Except for fifth position, which cannot drop 2 because it started at only 1. In case of a tie, the position that comes first is picked randomly. First, Brutus states his preferences: red, orange, yellow, purple, blue, green. Then Craw would states his: purple, blue, green, yellow, orange, red. Except that Craw is a meanie and adjusts his to make it unfair. But Brutus then adjusts his, and they go back and forth until neither can improve their score. The challenge is to figure out what the votes, standings, and satisfaction will be for Brutus and Craw will be when they finish, if they ever do.

Satisfaction\\[16pt] \begin{matrix} & 1st & 2nd & 3rd & 4th & 5th & 6th\\ \#1 & 5 & 3 & 2 & 1 & 0 & 0\\ \#2 & 4 & 4 & 2 & 1 & 0 & 0\\ \#3 & 3 & 3 & 3 & 1 & 0 & 0\\ \#4 & 2 & 2 & 2 & 2 & 0 & 0\\ \#5 & 1 & 1 & 1 & 1 & 1 & 0\\ \#6 & 0 & 0 & 0 & 0 & 0 & 0\\ \end{matrix}
Solution