Eleven liters of SAE 30 oil weighs 92 N. Calculate the oil's a. Specific weight b. Density Specific gravity
Eleven liters of SAE 30 oil weighs 92 N. Calculate the oil's a. Specific weight b. Density Specific gravity
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![### 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:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a2e60b5-531b-408d-ad87-91178dbcae31%2F0a4fbb1a-5f33-4c2c-8f74-043aaf8fe099%2Fgv56s7_processed.gif&w=3840&q=75)
Transcribed Image Text:### 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:
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