Let's go through each part of the question step by step.
a) What is the volume of the liquid?
The volume \( V \) of a cylinder can be calculated using the formula:
\[ V = \text{Base Area} \times \text{Height} \]
Given:
- Base Area \( A = 0.14 , \text{m}^2 \)
- Height \( h = 0.28 , \text{m} \)
Substituting the values:
\[ V = 0.14 , \text{m}^2 \times 0.28 , \text{m} = 0.0392 , \text{m}^3 \]
b) What is the mass of the liquid?
The mass \( m \) of the liquid can be calculated using the formula:
\[ m = \text{Density} \times \text{Volume} \]
Given:
- Density \( \rho = 4.8 \times 10^3 , \text{kg/m}^3 \)
Substituting the values:
\[ m = 4.8 \times 10^3 , \text{kg/m}^3 \times 0.0392 , \text{m}^3 \approx 188.736 , \text{kg} \]
c) What is the weight of the liquid?
The weight \( W \) of the liquid can be calculated using the formula:
\[ W = m \times g \]
where \( g \) is the acceleration due to gravity (approximated as \( 9.81 , \text{m/s}^2 \)).
Substituting the values:
\[ W = 188.736 , \text{kg} \times 9.81 , \text{m/s}^2 \approx 1,848.36 , \text{N} \]
d) What pressure does the liquid exert on the bottom of its container?
The pressure \( P \) exerted by the liquid at the bottom of the container can be calculated using the formula:
\[ P = \frac{W}{A} \]
where \( A \) is the area of the base.
Since we already have the weight \( W \) (approximately 1848.36 N) and the area \( A = 0.14 , \text{m}^2 \):
\[ P = \frac{1848.36 , \text{N}}{0.14 , \text{m}^2} \approx 13,202.57 , \text{Pa} , \text{(Pascals)} \]
In summary:
- a) Volume of the liquid: \( 0.0392 , \text{m}^3 \)
- b) Mass of the liquid: \( \approx 188.736 , \text{kg} \)
- c) Weight of the liquid: \( \approx 1,848.36 , \text{N} \)
- d) Pressure exerted on the bottom: \( \approx 13,202.57 , \text{Pa} \)
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