### Problem Statement:Eleven liters of SAE 30 oil weighs 92 N. Calculate the oil’s:a. Specific weight b. Density c. Specific gravity#### Explanation:- **Specific Weight (γ):** The specific weight is defined as the weight per unit volume. It can be calculated using the following formula: \[ \gamma = \frac{W}{V} \] where: - \( \gamma \) is the specific weight (N/m³) - \( W \) is the weight (N) - \( V \) is the volume (m³)- **Density (ρ):** The density of a substance is defined as its mass per unit volume. It can be calculated using the following formula: \[ \rho = \frac{m}{V} \] where: - \( \rho \) is the density (kg/m³) - \( m \) is the mass (kg) - \( V \) is the volume (m³) The mass can also be derived from weight using the relationship: \[ W = mg \] where: - \( g \) is the acceleration due to gravity (approximately 9.81 m/s²)- **Specific Gravity (SG):** The specific gravity is the ratio of the density of the substance to the density of water. It is a dimensionless quantity: \[ SG = \frac{\rho_{\text{oil}}}{\rho_{\text{water}}} \] where: - \( \rho_{\text{oil}} \) is the density of the oil - \( \rho_{\text{water}} \) is the density of water (approximately 1000 kg/m³ at 4°C)### Note:To solve the calculations, you will need to convert the volume from liters to cubic meters (1 liter = 0.001 cubic meters). ### Example Calculation:1. Convert the volume of oil from liters to cubic meters: \[ V = 11 \, \text{liters} \times 0.001 \, \frac{\text{m}^3}{\text{liter}} = 0.011 \, \text{m}^3 \]2. Calculate the specific weight:

Name: ### Problem Statement:Eleven liters of SAE 30 oil weighs 92 N. Calcul
Uploaded: 2024-03-20T06:56:55.000Z
Channel: Bartleby
Description: ### Problem Statement:Eleven liters of SAE 30 oil weighs 92 N. Calculate the oil’s:a. Specific weight b. Density c. Specific gravity#### Explanation:- **Specific Weight (γ):** The specific weight is defined as the weight per unit volume. It can