Sure! Let's go through each of the questions step by step and find the solutions.
### Question 12: Specific Gravity of Oil
Given:
- Density of oil = 60.3 lb/ft³
- Standard density of water = 62.4 lb/ft³
Formula:
The specific gravity (SG) is calculated as the ratio of the density of the substance to the density of water.
[tex]\[ \text{Specific gravity of oil} = \frac{\text{Density of oil}}{\text{Density of water}} \][/tex]
Solution:
[tex]\[ \text{Specific gravity of oil} = \frac{60.3 \text{ lb/ft}³}{62.4 \text{ lb/ft}³} \approx 0.966 \][/tex]
So, the specific gravity of the oil is approximately 0.966.
### Question 13: Specific Gravity of Lead
Given:
- Density of lead = 11,300 kg/m³
- Standard density of water = 1,000 kg/m³
Formula:
The specific gravity (SG) is calculated as the ratio of the density of the substance to the density of water.
[tex]\[ \text{Specific gravity of lead} = \frac{\text{Density of lead}}{\text{Density of water}} \][/tex]
Solution:
[tex]\[ \text{Specific gravity of lead} = \frac{11,300 \text{ kg/m}³}{1,000 \text{ kg/m}³} = 11.3 \][/tex]
So, the specific gravity of lead is 11.3.
### Question 14: Stress on a Man's Heel
Given:
- Weight of man = 200 lb
- Area of heel = 6.0 in²
Note:
To find stress, we need to convert the area from in² to ft².
1 inch² = 0.00694444 ft².
Convert area from in² to ft²:
[tex]\[ \text{Area in ft}² = 6.0 \text{ in}² \times 0.00694444 \text{ ft}²/\text{in}² \approx 0.04166664 \text{ ft}² \][/tex]
Formula:
Stress is calculated as the force (weight) divided by the area over which the force is applied.
[tex]\[ \text{Stress} = \frac{\text{Weight}}{\text{Area}} \][/tex]
Solution:
[tex]\[ \text{Stress} = \frac{200 \text{ lb}}{0.04166664 \text{ ft}²} \approx 4800.003 lb/ft² \][/tex]
So, the stress on the man's heel is approximately 4800.003 lb/ft².
### Summary
- The specific gravity of oil is approximately 0.966.
- The specific gravity of lead is 11.3.
- The stress exerted by a 200 lb man on a 6.0 in² heel is approximately 4800.003 lb/ft².
