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Scientific Enquiry - Observation and Measurement

Grade 7Cambridge (IGCSE)

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Scientific observations can be qualitative (describing qualities using the five senses) or quantitative (using numerical values and units, such as 25.0 cm25.0 \text{ cm}).

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Measurements must always include a magnitude and a unit. The SI (International System of Units) includes the meter (mm) for length, kilogram (kgkg) for mass, and second (ss) for time.

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Accuracy refers to how close a measurement is to the true value, while Precision refers to how close a series of measurements are to each other.

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Parallax error is a common mistake in measurement caused by viewing a scale at an angle. To avoid this, always read at eye level and perpendicular to the scale.

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When measuring liquid in a graduated cylinder, measurements should be taken from the bottom of the meniscus (the curve of the liquid surface).

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Independent variables are the factors we change, while Dependent variables are the factors we measure (e.g., measuring the change in temperature ΔT\Delta T over time tt).

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Derived units are units calculated from SI base units, such as area (m2m^2), volume (m3m^3), and density (kg/m3kg/m^3).

📐Formulae

Density(ρ)=Mass(m)Volume(V)\text{Density} (\rho) = \frac{\text{Mass} (m)}{\text{Volume} (V)}

Volume of a rectangular prism=l×w×h\text{Volume of a rectangular prism} = l \times w \times h

Average (Mean)=∑recorded valuesn\text{Average (Mean)} = \frac{\sum \text{recorded values}}{n}

Area of a rectangle=l×w\text{Area of a rectangle} = l \times w

💡Examples

Problem 1:

A student measures the mass of a rock as 150 g150\text{ g} and its volume as 30 cm330\text{ cm}^3. Calculate the density of the rock.

Solution:

ρ=150 g30 cm3=5 g/cm3\rho = \frac{150\text{ g}}{30\text{ cm}^3} = 5\text{ g/cm}^3

Explanation:

To find the density, divide the mass (mm) by the volume (VV). The resulting unit is g/cm3\text{g/cm}^3 because the mass was in grams and the volume was in cubic centimeters.

Problem 2:

A beaker contains 50 ml50\text{ ml} of water. After placing a marble inside, the water level rises to 62 ml62\text{ ml}. What is the volume of the marble?

Solution:

Vmarble=Vfinal−Vinitial=62 ml−50 ml=12 mlV_{marble} = V_{final} - V_{initial} = 62\text{ ml} - 50\text{ ml} = 12\text{ ml}

Explanation:

This is the displacement method used to measure the volume of irregular objects. Since 1 ml=1 cm31\text{ ml} = 1\text{ cm}^3, the volume is 12 cm312\text{ cm}^3.

Problem 3:

Convert a length of 2.5 meters2.5\text{ meters} into centimeterscentimeters.

Solution:

2.5 m×100=250 cm2.5\text{ m} \times 100 = 250\text{ cm}

Explanation:

There are 100 cm100\text{ cm} in 1 m1\text{ m}, so we multiply the value in meters by 10210^2 to get the value in centimeters.