An angler wants to use a spring-based balance to find the mass of a large salmon he has caught. The balance is graduated in kilograms instead of Newtons.
The balance is not calibrated correctly: with nothing hanging from it, the scale shows 1.5 kg1.5\text{ kg}1.5 kg.
Calculate the weight (in Newtons) of a 1.5 kg1.5\text{ kg}1.5 kg mass on Earth, where the gravitational field strength is g=9.8 N/kgg = 9.8\text{ N/kg}g=9.8 N/kg.
The angler decides to verify the accuracy of the balance. Lacking standard calibration weights, he places heavy canned soup tins into a light mesh bag and hangs it from the scale. Each tin has a nominal mass of 0.4 kg0.4\text{ kg}0.4 kg.
His observations are recorded in the table below:
| Number of tins | Scale reading (kg) |
|---|---|
| 0 | 1.5 |
| 1 | 1.9 |
| 2 | 2.3 |
| 3 | 2.7 |
| 4 | 3.8 |
| 5 | 3.5 |
| 6 | 3.9 |
Identify the anomalous data point in the table. Briefly explain why it is anomalous and how you would treat it when plotting a line of best fit.
State the relationship between the number of tins and the scale reading (excluding the anomalous point).
The angler remembers that six tins gave a scale reading of 3.9 kg3.9\text{ kg}3.9 kg. He calculates:
mass of six tins=6×0.4 kg=2.4 kg \text{mass of six tins} = 6 \times 0.4\text{ kg} = 2.4\text{ kg} mass of six tins=6×0.4 kg=2.4 kgand notes that:
3.9 kg−1.5 kg=2.4 kg 3.9\text{ kg} - 1.5\text{ kg} = 2.4\text{ kg} 3.9 kg−1.5 kg=2.4 kgHe concludes: "I can use this balance as normal! All I need to do is subtract 1.5 kg1.5\text{ kg}1.5 kg from each reading to get the correct mass."
He hangs his caught salmon from the balance. The reading on the scale is 6.1 kg6.1\text{ kg}6.1 kg. He concludes that his salmon must have a mass of exactly 4.6 kg4.6\text{ kg}4.6 kg.
Suggest three reasons why the angler's conclusions might be incorrect.
