In this challenge, the objective is to find the distance between 2 rods pulling elastics given that those elastics have equal heights. Now, because the elastics have different nominal lenghts and spring constants, they will only have equal lengths at a specific distance. This is because they must both be pulling with equal force. The force exerted by the two elastics is given by the first 2 equations below. They are equal, so we can solve them like a system of equations. We get that both are stretched to a length of 9 cm. The elastics are folded over twice, so their total length is thus only 4.5 centimetres. The combined force of the 2 elastics is 24 cmkg/s2. The weight resting upon the elastics will exert a downward force of 1 g * ≈9.807 m/s2 ≈ 9.807 mg/s2. In order for the elasitics to support it’s weight, the elastics must have an elasics must have this upwards force component. Let the angle below horizontal of the elastics be theta. θ ≈ asin(≈9.807/240), and the vertical distance between the rods is given by 4.5 * cos(asin(≈0.04086)). Using trigonometric identities, we simplify to the equation below. We get a distance of ≈8.992 centimetres.
\vec F_A=-{2kg\over s^2}(D-3cm)\\[8pt] \vec F_B=-{3kg\over s^2}(D-5cm)\\[16pt] \vec F_A=\vec F_B\\[8pt] {Dkg\over s^2}={9cmkg\over s^2}\\[8pt] D=9cm\\[32pt]F_A+F_B=24{cmkg\over s^2}\\[8pt] F_G\approx9.807{mg\over s^2}\\[16pt] x\approx4.5\cdot2\cdot\cos{\Bigg(\arcsin{\bigg({\approx9.807\over240}\bigg)}\Bigg)}\\[8pt] \approx\sqrt{1-\approx{0.04086}^2}\\[8pt] \approx8.992cmMonth: December 2019
Esoteric Elastics
This challenge is about elastics.
We begin with 2 elastics. Elastic A has nominal length 3 centimetres and spring constant 2 kilograms per square second. This notation seems unfamiliar but remember that we multiply by the displacement, in metres, and get a force, in mkg/s2. Elastic B has nominal length 5 centimetres and spring constant 3 kg/s2. The long elastic it threaded through the small one. The 2 ends of the long elastic are folded together. 2 identical rods are threaded through them. They are then pulled apart. The elastics are held in place at equal heights on the rods so they cannot move up or down. They are at equal heights and can rotate frictionlessly. The rods are then fastened to the surface of the earth some distance apart. The rods do not sway, bend, or in any way stop being vertical. A weight is placed halfway between the 2 rods, resting on the elastics. It weighs 1 gram. The weight does not cause any friction on the elastics. Elastics A and B are stretched to equal lengths. How far apart are the rods?
Solution to Curious Colonies
We wanted to find a way to position the colonies so that they all have the same area. It’s very interesting and fun to make cool pictures, models, and simulations of bacterial colony growth. But even just theorizing will do for this problem. The slowest colony, the Dotty bacteria colony, only grows at 1 centimetre a minute, but must still occupy 20% of the area. If the Dotty colony is with only the Linear bacteria, and opposite them, it will still fall short of 20%. Thus it is clearly impossible. You must tell Joana that the colonies cannot all start at the border and occupy equal areas. She must choose one.
Curious Colonies
This science challenge requires an understanding of bacterial growth.
You receive an internship at Joana the scientist’s bacteria laboratory. Remember Joana from Pure Water and Spring Forward? She’s studying some exquisite bacteria. No bacteria colonies can grow over others, but they just block each other’s progress. Joana is studying them on a circular dish with diameter 1 metre. Zigzag bacteria colonies’ radii increase by 4 centimetre a minute. Linear bacteria grow at 2 centimetres a minute. Checkerboard bacteria grow at 5 centimetres a minute. Dotty bacteria grow by 1 centimetre every minute, and Inscribed bacteria grow by 3 centimetres a minute. Joana wants to display the results by simultaneously starting one colony of each type on the border of the dish. After some time, all the bacteria colonies will reach an equilibrium. The challenge is to determine where to place the colonies so that they will all end with an equal portion of the dish.