Scientific Practices, Measurement & Data — Science

Apply it in a new setting

Problem 1

A student needs the density of an unknown metal sample. On a balance it reads 135 g. Dropped into a graduated cylinder, it raises the water level from 40.0 mL to 55.0 mL. What is the metal's density, in g/cm³? (1 mL = 1 cm³.)

Hints

Density = mass ÷ volume — but you must first find the volume from the water it pushes aside.

The displaced volume is the rise in the water level: 55.0 mL − 40.0 mL.

Divide the 135 g mass by that volume in cm³.

Answer and explanation

9 g/cm³

Volume by displacement = 55.0 mL − 40.0 mL = 15.0 mL = 15.0 cm³.

Density = mass ÷ volume = 135 g ÷ 15.0 cm³.

135 ÷ 15 = 9 g/cm³ (far denser than water, so it would sink).

Problem 2

A faucet drips steadily, losing 3 mL of water every 5 seconds. If it keeps dripping all day, how many liters of water are wasted in 24 hours?

Hints

First find the leak rate in mL per second, then find how many seconds are in a full day.

Seconds in a day = 24 × 60 × 60 = 86 400 s.

Multiply the rate by the seconds to get mL, then divide by 1000 to convert mL → L.

Answer and explanation

51.84 L

Rate = 3 mL ÷ 5 s = 0.6 mL/s.

One day = 24 × 60 × 60 = 86 400 s, so volume = 0.6 mL/s × 86 400 s = 51 840 mL.

Convert to liters: 51 840 mL ÷ 1000 = 51.84 L.

Problem 3

A student wants to test whether warmer water makes an antacid tablet dissolve faster. She has four cups. Which plan is a fair test that would give trustworthy evidence?

Use a different amount of water and a different-sized tablet in each cup, so every trial is unique
Change both the water temperature and the amount of stirring together, to save time
Keep the volume of water and the size of the tablet the same in every cup, and change only the water temperature
Compare one hot cup tested today against one cold cup she measured last week under conditions she no longer remembers

Hints

A fair test changes only ONE variable (the independent variable) and holds everything else constant.

The independent variable here is water temperature — everything else must be a controlled variable.

Look for the plan that keeps water amount, tablet size, and stirring identical, changing only temperature.

Answer and explanation

Keep the volume of water and the size of the tablet the same in every cup, and change only the water temperature

The independent variable is water temperature; a fair test changes only that and controls the rest.

Options that change several things at once (different water/tablet, or temperature + stirring) are confounded, and comparing against an unknown past trial has no valid control.

Only the plan that holds water volume and tablet size constant while varying temperature is a fair test — C.

Problem 4

A cart rolls at a steady speed while a motion sensor logs its position: | Time (s) | 0 | 2 | 4 | 6 | | Position (cm) | 0 | 30 | 60 | 90 | If the cart keeps this steady speed, what position (in cm) will it reach at t = 10 s?

Hints

Find the steady rate first: how many cm does the cart move each second?

The position rises 30 cm every 2 s, which is 15 cm every second.

Extend the pattern to t = 10 s: 15 cm/s × 10 s.

Answer and explanation

150 cm

The cart gains 30 cm every 2 s, so its speed is 30 ÷ 2 = 15 cm/s.

At a steady 15 cm/s, position = speed × time = 15 × 10.

15 × 10 = 150 cm.

Problem 5

To find a reliable value, a student times how long a ball takes to roll down a ramp, repeating it 5 times: 2.1 s, 2.3 s, 2.2 s, 2.2 s, and 3.5 s. On the 3.5 s run she saw the ball bump the wall, so she treats it as an anomaly and drops it. Using only the four consistent trials, what is the average time, in seconds?

Hints

Repeating trials and averaging cancels out random error — but a known mistake (the anomaly) is excluded first.

Average only the four trials: 2.1, 2.3, 2.2, and 2.2.

Add those four and divide by 4.

Answer and explanation

2.2 s

Drop the flagged anomaly (3.5 s) and keep the four consistent trials.

Sum = 2.1 + 2.3 + 2.2 + 2.2 = 8.8 s.

Average = 8.8 ÷ 4 = 2.2 s.

Problem 6

A student hangs masses on a spring and graphs the stretch: 100 g → 4 cm, 200 g → 8 cm, 300 g → 12 cm, and the points fall on a straight line through the origin. Reading the graph, how far (in cm) would a 250 g mass stretch the spring?

Hints

The straight line means the stretch is proportional to the mass — find the stretch per gram.

Every 100 g adds 4 cm, so each gram adds 0.04 cm.

250 g sits halfway between the 200 g (8 cm) and 300 g (12 cm) points.

Answer and explanation

10 cm

The line gives a constant rate: 4 cm per 100 g = 0.04 cm per gram.

Stretch at 250 g = 0.04 cm/g × 250 g (or halfway between 8 cm and 12 cm).

0.04 × 250 = 10 cm.

Problem 7

A scatter plot of many students shows that those who slept more hours the night before also tended to score higher on a quiz — the dots drift upward from left to right. Which conclusion does the graph best support?

Sleeping more definitely causes the higher quiz scores
There is no relationship between hours of sleep and quiz score
Scoring higher on the quiz must cause students to sleep more
The data show a positive relationship, but on their own they do not prove that more sleep causes higher scores

Hints

An upward drift in a scatter plot shows the two variables tend to rise together — a positive relationship.

Recall the rule: a relationship between two variables is not the same as one causing the other.

A hidden third factor (a studious routine, less screen time) could drive both.

Answer and explanation

The data show a positive relationship, but on their own they do not prove that more sleep causes higher scores

The upward-drifting dots show a positive relationship between sleep and score.

But correlation is not causation: the graph alone can't rule out reverse cause or a hidden third factor, so 'definitely causes' and 'must cause' overreach, while 'no relationship' ignores the clear trend.

The supportable reading is a positive relationship that does not by itself prove cause — D.

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