How to do calculations in science

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Example 1. Speed, distance, and time

Problem	How long does it take a bee flying at 20 m/s to cover a distance of 100 m?
List what you know	$$s = 20~\ce{m/s}, d = 100~\ce{m}, t = ?$$
Write out the equation	speed = distance / time, $$s=\frac{d}{t}$$
Substitute in your values	$$20=\frac{100}{t}$$
Rearrange if necessary	Multiply both sides by t. $$20t = 100$$ Divide both sides by 20. $$t = \frac{100}{20}$$
Calculate your answer	$$t = 5$$
Check and format the answer, showing units. Is your answer sensible?	$$t = 5~\ce{s}$$ (no requirements in the question for decimal places or significant figures). Sensible = yes.
Check by backwards calculation.	$$s=\frac{100}{5}=20~\ce{m/s}$$ Correct.

Example 2. Concentration, volume, and number of moles

Problem	How many moles are there in \(25~\ce{cm^3}\) of \(0.15~\ce{mol/dm^3}\) sodium hydroxide? Give your answer to 2 significant figures.
List what you know	$$\ce{Volume}, V = 25~\ce{cm^3} = \frac{25}{1000}~\ce{dm^3}$$; $$\ce{Concentration}, c = 0.15~\ce{mol/dm^3}$$; $$\ce{Number~of~moles}, n = ? $$
Write out the equation	concentration = number of moles/volume in dm3, $$c=\frac{n}{V}$$
Substitute in your values	$$0.15=\frac{n}{25/1000}$$
Rearrange if necessary	Multiply both sides by 25/1000. $$\frac{0.15 \times 25}{1000} = n $$
Calculate your answer	$$n = 3.75 \times 10^{-3}$$
Check and format the answer, showing units. Is your answer sensible?	(2 significant figures needed, so round to 3.8. Add units) $$\ce{Final~answer}, n = 3.8 \times 10^{-3}~\ce{mol}$$ Sensible = yes.
Check by backwards calculation.	$$c=\frac{n}{V} = \frac{3.75 \times 10^{-3}}{25/100} = 0.15$$ Correct.

Example 3. Energy, mass, specific heat capacity

Problem	What will be the change in temperature when 5000 J of thermal energy are transferred to a block of copper of mass 20g? The specific heat capacity of copper is 385 J/kg\(^{\circ}\)C.
List what you know	$$\ce{Energy}, E = 5000~\ce{J} $$ $$\ce{Mass}, m = 20~\ce{g}$$ $$\ce{Specific~heat~capacity}, c = 385~\ce{J/kg^{\circ}C} $$ Notice that the specific heat capacity is given in kg units, so we need to convert our mass into kg. $$20~\ce{g} = 20/1000~\ce{kg} = 0.02~\ce{kg}$$
Write out the equation	Energy = mass x specific heat capacity x change in temperature, $$E = mc\Delta\theta$$
Substitute in your values	$$5000 = 0.02 \times 385 \times \Delta\theta$$
Rearrange if necessary	Divide both sides by \(0.02 \times 385 \) $$\frac{5000}{0.02 \times 385} = \Delta\theta $$
Calculate your answer	$$\Delta{}\theta = \frac{5000}{0.02 \times 385} = 649.3506...$$
Check and format the answer, showing units. Is your answer sensible?	(The smallest number of significant figures in the numbers we are given is 2 in the mass of 20g, so we should round to 2 s.f. in our answer. Add units.) $$\ce{Final~answer}, n = 650~^{\circ}\ce{C}$$ Sensible = yes.
Check by backwards calculation.	$$E = mc\Delta\theta = 0.02 \times 385 \times 650 = 5005.$$ Correct.

Example 4. Kinetic energy, mass, velocity

Problem	A car of mass 1050 kg has a kinetic energy of 472,500 J. What is its velocity? Give your answer in m/s.
List what you know	$$\ce{Kinetic~energy}, E_k = 472,500~\ce{J} $$ $$\ce{Mass}, m = 1050~\ce{kg}$$
Write out the equation	$$\ce{Kinetic~energy} = \frac{1}{2} \times \ce{mass} \times \ce{velocity}^2$$ $$E_k = \frac{1}{2}mv^2$$
Substitute in your values	$$472,500 = \frac{1}{2} \times 1050 \times v^2$$
Rearrange if necessary	Multiply both sides by 2. $$2 \times 472,500 = 1050 \times v^2$$ Divide both sides by 1050. $$\frac{2 \times 472,500}{1050} = v^2$$ Take the square root of both sides. $$\sqrt{\frac{2 \times 472,500}{1050}} = v$$
Calculate your answer	$$v = \sqrt{\frac{2 \times 472,500}{1050}} = \sqrt{900} = 30$$
Check and format the answer, showing units. Is your answer sensible?	The answer is required in m/s. As the energy was in J, and the mass in kg, both standard SI units, our answer will automatically be in m/s. $$\ce{Final~answer}, v = 30~\ce{m/s}$$ Sensible = yes.
Check by backwards calculation.	$$E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 1050 \times 30^2 = 472,500.$$ Correct.

Example 5. A simple pendulum: period, length, and acceleration due to gravity.

Problem	A scientist observes a simple pendulum of length 4.00 m, and notes that the time taken to oscillate is 4.013 s. From these measurements, what is the acceleration due to gravity? Give your answer to an appropriate number of significant figures and give the units.
List what you know	$$\ce{Length}, l = 4.00~\ce{m}$$ $$\ce{Period}, T = 4.013~\ce{s}$$
Write out the equation	$$T = 2\pi \sqrt{\frac{l}{g}}$$
Substitute in your values	$$4.013 = 2\pi \sqrt{\frac{4.00}{g}}$$
Rearrange if necessary	Divide both sides by \(2\pi\). $$\frac{4.013}{2\pi} = \sqrt{\frac{4.00}{g}}$$ Square both sides. $$\left(\frac{4.013}{2\pi}\right)^2 = \frac{4.00}{g}$$ Let's get g on the top of the fraction. Multiply both sides by g. $$g \times \left(\frac{4.013}{2\pi}\right)^2 = 4.00$$ Multiply both sides by \(2\pi^2\). $$g \times 4.013^2 = 4.00 \times (2\pi)^2$$ Divide both sides by 4.013\(^2\). $$g = \frac{4.00 \times (2\pi)^2}{4.013^2}$$
Calculate your answer	$$g = \frac{4.00 \times (2\pi)^2}{4.013^2} = 9.805763366$$
Check and format the answer, showing units. Is your answer sensible?	The answer is an acceleration, which is a change in velocity (m/s) per unit time (/s), so has units m/s\(^2\). We are given 3 significant figures in the length supplied, so can only quote 3 s.f. in the answer. $$\ce{Final~answer}, g = 9.81~\ce{m/s}^2$$ Sensible = yes.
Check by backwards calculation.	$$T = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{4.00}{9.81}} = 4.012 \ce{s}$$ Correct.