AP Physics 1- 8.1 Internal Structure and Density- Study Notes- New Syllabus
AP Physics 1-8.1 Internal Structure and Density – Study Notes
AP Physics 1-8.1 Internal Structure and Density – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Internal Structure and Density of Fluids
Internal Structure and Density of Fluids
A fluid is a substance that can flow and does not resist shear stress. It includes both liquids and gases. Fluids have no fixed shape and take the shape of their container.
Key Properties of Fluids:
Density (\( \rho \)): Mass per unit volume of the fluid.
\( \rho = \dfrac{m}{V} \)
Where \( m \) = mass of fluid, \( V \) = volume.
Compressibility: Liquids are nearly incompressible, while gases are highly compressible.
Isotropy: At rest, fluids exert equal pressure in all directions. An ideal fluid is incompressible and has no viscosity.
Continuum Assumption: Fluids are treated as continuous matter even though they are made of discrete molecules.
Density Variations:
- Homogeneous fluids: Constant density throughout (e.g., pure water).
- Heterogeneous fluids: Density changes with location (e.g., atmosphere where air density decreases with altitude).
Units of Density:
- SI Unit: \( \text{kg/m}^3 \)
- Common unit: \( \text{g/cm}^3 \)
Pressure–Density Relationship:
In a static fluid under gravity, pressure increases with depth due to the weight of the fluid above:
\( P = P_0 + \rho g h \)
Where \( P_0 \) = pressure at surface, \( h \) = depth, \( g \) = gravitational acceleration.
Example :
Find the density of a liquid if 200 g of it occupies 250 mL.
▶️ Answer/Explanation
Mass \( m = 200 \, \text{g} = 0.2 \, \text{kg} \)
Volume \( V = 250 \, \text{mL} = 250 \times 10^{-6} \, \text{m}^3 \)
Density: \( \rho = \dfrac{m}{V} = \dfrac{0.2}{250 \times 10^{-6}} = 800 \, \text{kg/m}^3 \)
Answer: \( \rho = 800 \, \text{kg/m}^3 \)
Example :
At what depth in water ( \( \rho = 1000 \, \text{kg/m}^3 \) ) does the pressure increase by \( 2 \times 10^5 \, \text{Pa} \)?
▶️ Answer/Explanation
Using \( \Delta P = \rho g h \)
\( h = \dfrac{\Delta P}{\rho g} = \dfrac{2 \times 10^5}{1000 \times 9.8} \approx 20.4 \, \text{m} \)
Answer: Depth ≈ 20.4 m