Round 271

Date Released: July 17th, 2008.
Date Rewarded: July 24th, 2008.

Assume the world of Neopia is a spherical planet with a diameter of 1600 km with a uniform density of 5200 kilograms per cubic metre. If the moon of Kreludor is in a geostationary orbit around Neopia, and assuming a Neopian sidereal day is exactly 24 hours, how far is that geostationary orbit above the surface of Neopia?

Please submit your answer in kilometres, and round to the nearest kilometre. Submit only a number; any additional information may disqualify your entry.

Prize


Kreludan Defender Plushie

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This puzzle is about Geostationary orbit. You can find a full explanation about this topic in [1]. The main equation we need to focus on is:

r³ = G*M / ω²

where r is the distance between the Neopia's center of mass to the Kreludor moon's geostationary orbit, G = 6.674 28 * 10^-11 m³/(kg*s²) is the gravitational constant, M is the mass of the Neopia and ω is the angular speed of the moon.

The mass of the Neopia planet can be calculated by the following equation:

M = ρ * V = ρ * 4/3 * pi * R³

where ρ = 5200 kg/m³ is the density of Neopia, V is the volume of the planet, R = d/2 = 800 km = 800,000 m is the radius of the planet.

The angular speed of the Kredulor moon can be calculated by the following equation:

ω = (2 * pi) / 1 sidereal day = (2 * pi) / (24 * 60 * 60) = pi/43200 rad/s

Therefore, we have:

r³ = G * ρ * 4/3 * pi * R³ / (pi/43200)² = (86400)² * G * ρ * R³ / (3*pi)

Replace numbers into the above equation, then take cubic root, we can get r = 5201692.7355. The answer is the distance from the Neopia planet surface to the orbit. Therefore, it should be:

distance = (r - R)/1000 = 4401.69

We use 1000 factor in the above equation because we want to get in kilometers. Round the distance and we get 4402 as the answer.

Results: 526 people guessed the correct answer earning themselves 3803 NP each.

References:
[1] Geostationary orbit

Round 260

Date Released: May 1st, 2008.
Date Rewarded: May 8th, 2008.

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

Suppose you have a very long steel bar, with a regular pentagonal cross-section where each side of the pentagon is 2 cm in length. This bar will be suspended vertically over a very deep chasm. At the bottom end of the steel bar, there is a spherical steel weight with a diameter of 50 cm. Going up the bar, at every 3 metre increment, there is a spherical steel weight with a diameter of 30cm.

If the density of the steel is exactly 7.8 g/cm³, the steel can withstand a tensile stress of 400 MPa, and gravity is 9.81m/s², what is the maximum possible length, in metres, that the bar can extend without breaking? Round down to the nearest metre, submit only a number (with no additional or extraneous information), and disregard any deformation of the bar or any other forces acting on the bar.

Prize


Steel Negg

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Using the stress equation [1]:

rho = F/A

where rho = 400 MPa = 400,000,000 kg/m·s² is the tensile stress that steel can withstand, F is the force that the steel has to withstand and A is the crossed area of the steel bar.

The crossed area is a regular pentagon can be calculated by the following equation [2]:

A = 5/4 t² tan(54°) = 0.000688190960236 m²

where t = 2 cm = 0.02 m is the side of the pentagon.

Now, we can calculate the force that the steel bar can withstand:

F = rho * A = rho * 5/4 t² tan(54°)

Using Newton's gravity law [3], we have:

F = m * g

Therefore, the mass that the steel bar can widthstand is:

m = F /g = rho * 5/4 t² tan(54°) / g = 28060.793485650833 kg

where g = 9.81 m/s² is the gravitational acceleration.

Don't round anything yet. The mass is contributed by 3 components: the big sphere m₁, small spheres m₂, and the steel bar m₃. Since all of them are steel, they have the same density. We have:

m₁ + int(h/3) * m₂ + m₃ = m
<=>[V₁ + int(h/3) * V₂ + A * h] * d = m

where h is the length of the steel bar, int is the floor function (every 3 meter, we have one small sphere), d = 7.8 g/cm³ = 7,800 kg/m³ is the density of steel, V₁, V₂ are the volumes of big and small spheres. The volume of the steel bar is the product of the crossed area with the length of the bar.

We can calculate V₁, V₂ by the sphere volume formula [4]:

V₁ = 4/3 * pi * (r₁)³ = 0.065449846949787 m³
V₂ = 4/3 * pi * (r₂)³ = 0.014137166941154 m³

where the radius r₁ = d₁/2 = 25 cm = 0.25 m, r₂ = d₂/2 = 15 cm = 0.15 m. Don't round both volume either. We already know A, so the equation now should be:

V₁ + int(h/3) * V₂ + A * h = m/d
<=>int(h/3) * V₂ + A * h = m/d - V₁

We know the right hand side of the equation but the left hand side is a little bit tricky. So, we can do a method of trial-and-error to test. Solve h by the following equation:

h/3 * V₂ + A * h = m/d - V₁

<=>(V₂/3 + A) * h = m/d - V₁
<=>h = (m/d - V₁)/(V₂/3 + A) = 654.0200901109209 m

Now round h down to an integer. We then have h = 654 m. Next, test h into the equation int(h/3)* V₂ + A * h = m/d - V₁. If the left hand side is smaller than the right hand side, the steel bar can withstand the gravitational forces. The largest h that is satisfied that the left hand side is smaller than the right hand side will be the answer.

Using h = 654 m, the left hand side: int(h/3)* V₂ + A * h = 3.5320. The right hand side: m/d - V₁ = 3.5321 (good). Now increase h by 1 (h = 655 m) to see if the steel bar can still withstand. The left hand side: int(h/3)* V₂ + A * h = 3.5327. The right hand side: m/d - V₁ = 3.5321. (not good). Therefore, the answer is h = 654 m.

Results: 986 people guessed the correct answer earning themselves 2029 NP each.

References:
[1] Stress
[2] Regular pentagon area
[3] Newton's gravity law
[4] Sphere volume

Round 258

Date Released: April 17th, 2008.
Date Rewarded: April 24th, 2008.

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Answer this question, and you may pass," she said.

Suppose you have a perfectly spherical water tank with an inside diameter of 8.6 metres. If the drain at the bottom of the tank can't handle a hydrostatic pressure of more than 50 kilopascals, what is the maximum volume of water, in litres, that can be contained in the tank? Assume that gravitational acceleration is exactly 9.81 m/s². Please round to the nearest 10 litre increment, and please submit only a number for your answer. (For example, if you calculate the answer to be 16277 litres, submit 16280 as your answer)

Prize


Underwater Tour

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First off, we need to find the maximum height of the water tank that the drain can handle using hydrostatic pressure equation [1]:

P = rho * g * h <=> h = P / (rho * g) = 5.096839959225281 m

where P = 50k Pa = 50,000 kg/m·s² is the maximum hydrostatic pressure, rho = 1000 kg/m³ is density of water, g = 9.81 m/s² is the gravitational acceleration. Remember that do not round h yet.

Then, we can start to calculate the partial volume of the sphere by the following formula (explanation of the formula is at the end of the solution):

V = 1000 * pi * h² * (r - h/3) = 212276.11

where r = d/2 = 4.3 m, is the radius of the spherical water tank, h is the above maximum height.

We know that: 1 m³ = 1000 dm³ = 1000 liters. We have a 1000 factor in the above formula because the answer should be in liters while we calculated in cubic meters. Now perform rounding as the requirements and we get V = 212,280 liters.

Results: 462 people guessed the correct answer earning themselves 4330 NP each.

If you are interested to see how the partial volume of a sphere is derived, you can continue to read. Otherwise stop here.


We can consider the sphere with center at origin (above picture). We cut the partial sphere into small slices. Each slice is a circle, perpendicular to the vertical direction. Using Pythagorean theorem [2], we can calculate area of each circle at a certain height x:
Then we do integration on the whole vertical direction to add all area of the slices to get the volume. Since we choose the origin at the center of the sphere, the limit of the integration should be from -r to h-r

References:
[1] Hydrostatic pressure
[2] Pythagorean theorem

Round 24

Date released: September 12th, 2001.
Date rewarded: October 5th, 2001.

Pawkeet and Mirgle have had their game of Deckball rudely interrupted by a filthy braggart pirate. In an attempt to ruin their fun he steals their deckball, loads it into his cannon and fires it off into the distance.

Assuming that the value of gravity on Neopia is 9.8m/s², and that the initial launch velocity of the ball is 100 meters per second, and also assuming that the ball lands on a rocky outcrop at exactly the same altitude as the cannon...

How many seconds is the deckball in the air?
(rounding down to the nearest second)

Prize
Unknown

Click to show/hide solution

The information in the problem is not enough to solve. If we want to get the answer that TNT gave, we have to assume the given initial velocity is the part in vertical direction. That means:

v_y = 100 m/s

According to the trajectory of a projectile equations [1], the time for the deckball to be in the air is:

t = 2vsinθ / g = 2v_y / g = 20.41 s

where θ is the angle of shooting, and g = 9.8m/s² is the Neopia gravity acceleration. After rounding, the answer is 20 seconds.

References:
[1] Trajectory of a projectile