Solid Copper Has A Density Of 8960 Kg/M³; What Is The Mass?
Solid Copper Has A Density Of 8960 Kgm3 What Is The Mass Of A Cub
1. Solid Copper has a density of 8960 Kg/m3. What is the mass of a cube of copper with side length 0.1 m? 2. Liquid mercury has a density of 13,534 Kg/m3. What is the pressure difference (gauge pressure) between a point on the surface and one point 1 m deep inside a tank of mercury? 3. Under the assumption that the density of the air does not change with the altitude, using equation P - P0 = ρgh, the air density ρ = 1 Kg/m3, P0 = 0 Pa outside earth atmosphere, P = 1×105 Pa at the earth surface, calculate the height h of the earth atmosphere. 4. A hydraulic pressure needs to lift a 3,000 Kg load (against gravity) by using an input force of 1,000 N. All frictions are negligible and the area of the platform under the load is 10 m2. Calculate the (maximum) area of the section of the pump where the input force is applied. 5. The following picture depicts a solid object completely submerged into a homogeneous fluid. (i) Is the pressure at point (A) higher, as high as, or lower than pressure at point (B)? (ii) Is the pressure at point (A) higher, as high as, or lower than pressure at point (C)?
Paper For Above instruction
The given set of physics problems offers an insightful exploration into material properties, fluid mechanics, and pressure principles, fundamental concepts in physics that have extensive applications in engineering and environmental sciences. This paper systematically addresses each problem, utilizing theoretical frameworks and mathematical calculations to arrive at comprehensive solutions.
1. Mass of a Copper Cube
Given the density (ρ) of copper as 8960 kg/m3 and the side length (a) of the cube as 0.1 meters, the volume (V) of the cube is calculated using the formula V = a3. Substituting the values yields V = (0.1)3 = 0.001 m3. The mass (m) can then be obtained by multiplying the volume by the density: m = ρV = 8960 × 0.001 = 8.96 kg. This straightforward calculation exemplifies how material density directly influences mass, crucial in designing components where weight is a concern.
2. Pressure Difference in Mercury
The pressure differential at a depth h in a fluid is determined by the hydrostatic pressure equation ΔP = ρgh. With the density of mercury ρ = 13,534 kg/m3, gravitational acceleration g ≈ 9.81 m/s2, and depth h = 1 m, the pressure difference becomes ΔP = 13,534 × 9.81 × 1 ≈ 132,676 Pa. This pressure difference, often termed gauge pressure, is significant in designing hydraulic systems, as it influences the structural integrity and fluid handling requirements of tanks and piping.
3. Height of Earth's Atmosphere
Using the barometric formula P - P0 = ρgh, with P = 1×105 Pa, P0 = 0 Pa, and ρ = 1 kg/m3, the height (h) can be calculated as h = (P - P0)/(ρg) = 1×105 / (1 × 9.81) ≈ 10,193 meters. This estimation aligns with the approximate height of the lower atmosphere, illustrating the application of hydrostatic principles in atmospheric science and climate modeling.
4. Hydraulic Lift: Calculating Area
Applying Pascal's principle, the force (F2) exerted by the fluid on the load is related to the input force (F1) and their respective areas (A1 and A2): F2 / A2 = F1 / A1. Rearranged to find A1: A1 = (F1 × A2) / F2. Substituting F1 = 1000 N, A2 = 10 m2, and F2 = m × g = 3000 × 9.81 = 29,430 N, results in A1 = (1000 × 10) / 29430 ≈ 0.34 m2. This calculation demonstrates the capacity constraints and efficiency considerations in hydraulic systems used for lifting heavy loads.
5. Pressure Comparison in a Fluid
In a fluid at rest, pressure increases with depth due to the weight of the overlying fluid. (i) At points A and B, assuming B is deeper than A, the pressure at A is higher or equal depending on their relative depths; specifically, pressure at A is lower than at B if B is below A. (ii) Comparing pressure at A and C, if C is located at an intermediate depth, the pressure at A and C depends on their positions; if C is deeper than A, the pressure at C exceeds that at A. These insights align with Pascal's law and are essential in understanding fluid statics and designing pressurized systems.
Conclusion
This comprehensive analysis illustrates the application of fundamental physics principles to practical problems involving material density, fluid pressure, and hydraulic systems. The calculations underscore the importance of these principles in engineering design, environmental modeling, and industrial applications, emphasizing the interconnectedness of physical laws and real-world systems.
References
- Serway, R. A., & Jewett, J. W. (2014). Physics for Scientists and Engineers with Modern Physics (9th ed.). Brooks Cole.
- Tipler, P. A., & Mosca, G. (2007). Physics for Scientists and Engineers. W. H. Freeman.
- Giancoli, D. C. (2013). Physics: Principles with Applications (7th ed.). Pearson Education.
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