﻿ Maths of Planet Earth | Limitless Applications

Classroom Resources

Published on November 1st, 2012 | by Simi

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Weather wrangling

1. 8.9°C, 5.4°C, 1.9°C
2. 34.7°C, 38.4°C, 42.1°C
3. 12.6 m/s, plot will have 4 parallel lines
4. 76%, plot will have 4 non-parallel lines

Crazy cacti!

Freeway fastlanes

Possible Possums

When will the greenhouse power a green house?

Power from the sun

1) 800/cos(30 deg) = 924W, slope 30 degrees from the horizontal facing northwards

2) 800 x cos(10 deg)/cos(30 deg) = 910W – not much difference from Q1 since cos(10 deg) is closer to 1 than you might think!

3) 300 x sin(70 deg) x sin(20 deg)/cos(70 deg) = 282W – please check 🙂

Close encounters

1. $\sqrt{358^2 – 277^2}$ x 100 x 2/7.8/60 ~ 96.9 minutes

2. 277 x tan(70 deg) x 100 x 2/7.8/60 ~ 144.7 minutes

Oxygen in the earth

1. Total mass of oxygen = 30.1% of 5.98 x 1024 kg ~ 1.80 x 1024 kg = 1.80 x 1027 g

2. Oxygen volume = mass / density = 1.79998 x 1027 g / 1.43 x 10-3 g / cm3 ~ 1.26 x 1030 cm3 = 1.26 x 1015 km3

3. Volume of earth = 4/3 pi r3 = 4/3 pi (6356.8)3 ~ 1.08 x 1012 km3

4. Number of earths’ volume of oxygen = 1.26 x 1015 km3 / 1.08 x 1012 km3 ~ 1.17 x 103 = 1170 earths.

Record pollution levels in Beijing

Evaporation for dams in the Toowoomba area

1 a) We assume 1 m3 of water contains 1000L, so 1000 m3 corresponds to 1ML. human consumption: 33.3 x 7 = 233.1 ML evaporation: 7 x 0.0065m x 170 x 1000 x 1000 = 7,735,000 m3 = 7,735 ML
Total: 7,968.1ML, fraction due to human consumption: 233.1/7968.1 = 2.9%

1 b) 0.013 x 170 x 1000 x 1000 = 2,210,000 m3 = 2,210 ML. This would fit in a cube of side length 130m

2. Area of pool is 9.2 x 4.4 = 40.48 m2, so on average 40.48 x 0.0089 x 1000 = 360.3 litres evaporates per day. Over 7 days 360.3 x 7 = 2521.9 L is lost, so it will take 2521.9/17 = 148.4 minutes to fill up – invest in a pool blanket!

Dust storm

Crisis call on climate

Can our family car really be emitting tonnes of greenhouse gases each year? Surely not!

Pass the small tweezers!

Mathematics of the mission to the edge of space

a) During freefall he fell 36529m in 4 minutes 20 seconds, or 260 seconds. Hence his average speed of descent was 36529/260 ≈ 140.50 m/s.

After freefall he was 39045 – 36529 = 2516m above sea level. Hence after freefall he fell 2516 – 1363 = 1153m. This took place in 4 minutes 43 seconds (9 minutes 3 seconds minus 4 minutes 20 seconds). This is equal to 283 seconds so his average speed of descent after freefall was 1153/283 = 4.07 m/s.

b) To find his acceleration in m/s2 we first convert his maximum speed to m/s: 1342.8/3.6 = 373m/s. Hence his acceleration was 373/42 = 8.88m/s2, or 0.91 times the standard acceleration due to gravity.

c) Assuming constant acceleration, his average speed was half his maximum speed, or 373/2 = 186.5 m/s. Hence after 42 seconds he descended 186.5 x 42 =7833 m.

Extension question: The graph shows a possible speed vs time graph.

It was made with the assumption that when the parachute was deployed he was travelling at a terminal velocity of approximately 200 km/h or 55 m/s. Note that many solutions are possible but a shape similar to this is required to ensure that the area under the graph matches the descent data.

Piles of Plankton

a. 10,000 kilograms of plankton
b. 1.05263 * 10^9 cubic metres of seawater.

It’s a small world

Diameter of a proton (as measured by charge radius) (~1.7 x 10^-15 m)

~1 order of magnitude to

Diameter of a uranium nucleus (~1.2 x 10^-14 m)

~4 orders of magnitude to

Diameter of a carbon atom (1.4 x 10^-10 m)

~5 orders of magnitude to

Diameter of a white blood cell (~1.2 x 10^-5 m)

~3 orders of magnitude to

Diameter of Australian 5c coin (1.9 x 10^-2 m)

~2 orders of magnitude to

Height of an adult human (~1.5 – 1.8 m)

It’s a big world

Height of an adult human (~1.5 – 1.8 m)

~2 orders of magnitude to

Length of the MCG field (1.7 x 10^2 m)

~1 order of magnitude to

Height of Mount Kosciuszko above sea level (2.228 x 10^3 m)

~3 orders of magnitude to

Distance from Melbourne to Perth (as the crow flies) (2.7 x 10^6 m)

~1 order of magnitude to

Diameter of the earth (1.3 x 10^7 m)

~1 orders of magnitude to

Diameter of the sun (1.4 x 10^9 m)

~2 orders of magnitude to

Distance from earth to sun (1.5 x 10^11 m)

~10 orders of magnitude to

Diameter of the Milky Way Galaxy (1 x 10^21 m)

Restriction enzymes and DNA palindromes

(a) 256, 1024, 65536. With 4 choices (A,C,G,T) for the nucleotide in
each position, there are 4^4 = 256 possible sequences of length 4. Of
length 5 there are 4^5 = 1024. Of length 8, 4^8 = 65536.

(b) 16, 64, 256. In a mirror palindromic sequence of length 4, the
opposite pairs of nucleotides must be equal; so there are 4 choices
for the nucleotides in first and second positions, and then the
sequence is determined. So there are 4^2 = 16 such sequences of length
4. With length 5, the first 3 choices of nucleotides determine the
sequence, so there are 4^3 = 64 such sequences. Of length 8 there are
4^4 = 256.

(c) 16, 0, 256. In an inverted repeat palindromic of length 4
Similarly to the mirror palindromic case, the first and second
nucleotides may be freely chosen and then the last two are determined
— the fourth must be complementary to the first, and the third to
the second. So there are 4^2 = 16 such sequences of length 4. Of
length 5 there are no such sequences, as the centre nucleotide would
have to be complementary to itself, which is impossible. With length
8, the first 4 nucleotides may be chosen freely, and the last 4 are
then determined, giving 4^4 = 256 sequences.

DNA

a) 6 positions, 3 choices for each position resulting in 18 new sequences
b) 4 new sequences: AACCTGG, ACCCTGG, ACCTTGG, ACCTGGG
c) 4 again: CCTGG, ACTGG, ACCGG, ACCTG
d) 22: we have the 4 found in b) and then those in which the new letter is different from its neighbour(s): 3 + 2 + 3 + 2 + 2 + 3 + 3 = 18 corresponding to XACCTGG, AXCCTGG, ACXCTGG, ACCXTGG, ACCTXGG, ACCTGXG, ACCTGGX.

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