Which tools would be use to find an irregularly shaped object’s mass and volume?
1 answer:
You might be interested in
Answer:
The depth to which the hydrometer sinks is approximately 24.07 cm
Explanation:
The given parameters are;
The diameter of the hydrometer tube, d = 2.3 cm
The mass of the content of the tube, m = 80 g
The density of the liquid in which the tube floats, ρ = 800 kg/m³
By Archimedes' principle, the up thrust (buoyancy) force acting on the hydrometer = The weight of the displaced liquid
When the hydrometer floats, the up-thrust is equal to the weight of the hydrometer which by Archimedes' principle, is equal to the weight of the volume of the liquid displaced by the hydrometer
Therefore;
The weight of the liquid displaced = The weight of the hydrometer, W = m·g
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
∴ W = 80 g × g
The volume of the liquid that has a mass of 80 g (0.08 kg), V = m/ρ
V = 0.08 kg/(800 kg/m³) = 0.0001 m³ = 0.0001 m³ × 1 × 10⁶ cm³/m³ = 100 cm³
The volume of the liquid displaced = 100 cm³ = The volume of the hydrometer submerged,
= A × h
Where;
A = The cross-sectional area of the tube = π·d²/4
h = The depth to which the hydrometer sinks
h = /A
∴ h = 100 cm³/( π × 2.3²/4 cm²) ≈ 24.07 cm
The depth to which the tube sinks, h ≈ 24.07 cm.
The gravitational potential energy
gpe = mgh
The missing diagram is in the attachments.
Answer: X: positive Y: positive
Explanation: Electric field is a vector quantity, which means it can be represented by a vector arrow: the arrow points in the direction of electric field and its length represents the magnitude at a given location. There are another representation of the electric field called electric field lines, <u>in which the line points away from a positively charged source and towards a negatively charged source</u>. This occurs because it follows a pattern, where the lines points in the direction that a positive test charge would have if it is accelerating on the line.
Analyzing the diagram, it can be observed that the lines are pointing away from both of the charged objects. Therefore, both X and Y are <u>positively charged</u>.
