**Published on** November 1st, 2012 |
*by Simi*

# ANSWERS

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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

Open CDF interactive for Weather wrangling

**Crazy cacti!**

Answers coming soon

**Freeway fastlanes**

Answers coming soon

**Possible Possums**

Answers coming soon

**When will the greenhouse power a green house?**

Answers coming soon

**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 10^{24} kg ~ 1.80 x 10^{24} kg = 1.80 x 10^{27} g

2. Oxygen volume = mass / density = 1.79998 x 10^{27} g / 1.43 x 10^{-3} g / cm^{3} ~ 1.26 x 10^{30} cm^{3} = 1.26 x 10^{15} km^{3}

3. Volume of earth = 4/3 pi r^{3} = 4/3 pi (6356.8)^{3} ~ 1.08 x 10^{12} km^{3}

4. Number of earths’ volume of oxygen = 1.26 x 10^{15} km^{3} / 1.08 x 10^{12} km^{3} ~ 1.17 x 10^{3} = 1170 earths.

**Record pollution levels in Beijing**

Answers coming soon

**Evaporation for dams in the Toowoomba area**

1 a) We assume 1 m^{3} of water contains 1000L, so 1000 m^{3} 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 m^{3} = 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 m^{3} = 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 m^{2}, 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**

Answers coming soon

**Crisis call on climate**

Answers coming soon

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

Answers coming soon

**Pass the small tweezers!**

Answers coming soon

**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.

**a.** 10,000 kilograms of plankton

**b.** 1.05263 * 10^9 cubic metres of seawater.

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)

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.

**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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