The pressure on the sample is 4.375 × 10⁷ Pa
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Further explanation
Let's recall Hydrostatic Pressure formula as follows:
\(\boxed{ P = \rho g h}\)
\(\boxed{ P = F \div A}\)
where:
P = hydrosatic pressure ( Pa )
ρ = density of fluid ( kg/m³ )
g = gravitational acceleration ( m/s² )
h = height of a column of liquid ( m )
F = force ( N )
A = cross-sectional area ( m² )
Let us now tackle the problem!
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Given:
diameter of large cylinder = D₁ = 10.0 cm
diameter of small cylinder = D₂ = 2.0 cm
area of the sample = A = 4.0 cm² = 4.0 × 10⁻⁴ m²
magnitude of force = F = 350 N
Asked:
pressure on the sample = P = ?
Solution:
Firstly , we will find the force acting on the small cylinder:
\(\Sigma \tau = 0\)
\(F(2L) - F_2(L) = 0\)
\(2FL = F_2L\)
\(F_2 = 2F\)
\(F_2 = 2(350)\)
\(\boxed{F_2 = 700 \texttt{ N}}\)
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Next, we could find the force acting on the large cylinder:
\(P_1 = P_2\)
\(F_1 \div A_1 = F_2 \div A_2\)
\(F_1 = \frac{A_2}{A_1} F_2\)
\(F_1 = \frac{\frac{1}{4} \pi (D_2)^2}{\frac{1}{4} \pi (D_1)^2} F_2\)
\(F_1 = (\frac{D_2}{D_1})^2 F_2\)
\(F_1 = (\frac{10.0}{2.0})^2 \times 700\)
\(\boxed{F_1 = 17500 \texttt{ N}}\)
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Finally, we could calculate the pressure on the sample:
\(P = F_1 \div A\)
\(P = 17500 \div (4 \times 10^{-4})\)
\(\boxed{P = 4.375 \times 10^7 \texttt{ Pa}}\)
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Learn more
- Buoyant Force : https://brainly.com/question/13922022
- Kinetic Energy : https://brainly.com/question/692781
- Volume of Gas : https://brainly.com/question/12893622
- Impulse : https://brainly.com/question/12855855
- Gravity : https://brainly.com/question/1724648
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Answer details
Grade: High School
Subject: Physics
Chapter: Pressure
