A Solid Block With Mass 4 Kg Floats On The Surface Of A Tan
A Solid Block With Mass 04 Kg Floats On The Surface Of A Tank Of W
A solid block with mass 0.4 Kg floats on the surface of a tank of water (density of water, ρwater = 1000 Kg/m3); the part of the solid submerged in the water is one third of its total volume. (i) Calculate the density of the solid. (ii) Calculate the total volume of the solid. An object has a weight of 24 N in air and 22 N when fully submerged in water (density of water, ρwater = 1000 Kg/m3). (i) Calculate the density of that object. (ii) Calculate the mass of that object (take g = 10 m/s2). (iii) Calculate the buoyancy force on that object when it is fully submerged in water. (iv) Calculate the buoyancy force on that object when it is half submerged in water (i.e., when half of its volume is under water). An ideal fluid flows at 2 m/s in a cylindrical pipe with diameter of 0.4 m. (i) If the pipe splits into 4 branches, each with 0.1 m diameter, what will be the speed of the fluid in each branch? (ii) If the pipe splits into 4 branches, all with the same diameter d, for what value of d will the speed of the fluid in each branch remain equal to 2 m/s? A mass of 0.2 Kg is hung to a spring. When displaced by 0.04 m, the mass oscillates with a frequency of 50 Hz. (i) Calculate the elastic constant of the spring. (ii) Calculate the time it will take the mass to complete 5 full oscillations. The picture below (bird view) depicts a spinning ball that travels in the direction of the arrow, through the air. Draw the trajectory the ball will follow. Thoughts on Social Work Knowledge Development Activities within a Quantitative Framework Grogan-Kaylor, Andrew;Delva, Jorge Social Work; Oct 2008; 53, 4; ProQuest Central pg. 293
Paper For Above instruction
The study of buoyancy, fluid dynamics, oscillatory motion, and projectile trajectory forms fundamental aspects of physics, providing insights into natural phenomena and engineering applications. This paper explores various problems related to these topics, exemplifying theoretical principles through practical calculations, with particular emphasis on buoyant forces, fluid flow in pipes, harmonic motion, and projectile motion trajectories within the context of physical laws.
Analysis of Buoyancy and Density of Floating Solid
Initially, we examine a solid block of mass 0.4 kg floating on water, with one-third of its volume submerged. The principle of buoyancy, governed by Archimedes' principle, states that the buoyant force equals the weight of displaced fluid. The density of the water (ρwater) is given as 1000 Kg/m3. Given that the submerged volume is one-third of the total volume (Vsubmerged = 1/3 Vtotal), we can infer the density of the solid (ρsolid) using the equilibrium condition:
\[
\rho_{solid} = \frac{mass}{V_{total}} = \frac{m}{V_{total}}
\]
Since the weight of the block (W) is 0.4 kg × g (assuming g = 10 m/s2), the weight is 4 N. The buoyant force when floating balances the weight:
\[
\text{Buoyant force} = \rho_{water} \times V_{submerged} \times g = W
\]
Substituting Vsubmerged:
\[
\rho_{water} \times \frac{1}{3} V_{total} \times g = 0.4 \times 10
\]
\[
1000 \times \frac{1}{3} V_{total} \times 10 = 4
\]
\[
\frac{1000 \times 10}{3} V_{total} = 4
\]
\[
V_{total} = \frac{4 \times 3}{1000 \times 10} = \frac{12}{10,000} = 0.0012\, \text{m}^3
\]
The density of the solid is then:
\[
\rho_{solid} = \frac{0.4}{V_{total}} = \frac{0.4}{0.0012} \approx 333.33\, \text{kg/m}^3
\]
This indicates that the solid is less dense than water, as expected for a floating object. The total volume of the solid is approximately 0.0012 m3.
Density and Buoyant Forces on a Submerged Object
Considering an object with a weight of 24 N in air and 22 N when submerged, we analyze its density and mass. The loss in apparent weight when submerged is due to the buoyant force:
\[
\text{Buoyant force} = \text{Weight in air} - \text{Weight in water} = 24\,N - 22\,N = 2\,N
\]
Since the buoyant force equals the weight of displaced water:
\[
F_b = \rho_{water} \times V_{object} \times g
\]
\[
2 = 1000 \times V_{object} \times 10
\]
\[
V_{object} = \frac{2}{10,000} = 0.0002\, \text{m}^3
\]
The mass of the object:
\[
m = \frac{W}{g} = \frac{24}{10} = 2.4\, \text{kg}
\]
The density of the object:
\[
\rho_{object} = \frac{m}{V_{object}} = \frac{2.4}{0.0002} = 12,000\, \text{kg/m}^3
\]
When half of the object is submerged, the buoyant force becomes:
\[
F_b = \rho_{water} \times V_{half} \times g = \rho_{water} \times \frac{V_{total}}{2} \times g
\]
\[
F_b = 1000 \times \frac{0.0002}{2} \times 10 = 1000 \times 0.0001 \times 10 = 1\,N
\]
This shows that the buoyant force reduces proportionally as the submerged volume decreases, reflecting the linear relationship between submerged volume and buoyant force in Archimedes' principle.
Fluid Flow in Pipes and Area Considerations
In the case of fluid flowing in a cylindrical pipe of diameter 0.4 m at 2 m/s, the volumetric flow rate (Q) is:
\[
Q = A \times v = \frac{\pi}{4} D^2 \times v
\]
Calculating the cross-sectional area:
\[
A = \frac{\pi}{4} \times 0.4^2 = \frac{\pi}{4} \times 0.16 \approx 0.1257\, \text{m}^2
\]
The flow rate:
\[
Q = 0.1257 \times 2 \approx 0.2514\, \text{m}^3/\text{s}
\]
When the pipe splits into four branches, each with diameter dbranch = 0.1 m, the area of each branch:
\[
A_{branch} = \frac{\pi}{4} \times 0.1^2 = 0.00785\, \text{m}^2
\]
Since the total flow must be conserved:
\[
Q_{total} = 4 \times A_{branch} \times v_{branch}
\]
Solving for the velocity in each branch:
\[
v_{branch} = \frac{Q}{4 \times A_{branch}} = \frac{0.2514}{4 \times 0.00785} \approx 8\, \text{m/s}
\]
Thus, the fluid accelerates significantly in smaller branches. To keep the velocity at 2 m/s in each branch, the diameter dbranch should satisfy:
\[
Q = A_{branch} \times v_{desired}
\]
\[
A_{branch} = \frac{Q}{v_{desired}} = \frac{0.2514}{2} = 0.1257\, \text{m}^2
\]
Corresponding diameter:
\[
d = \sqrt{\frac{4A_{branch}}{\pi}} \approx \sqrt{\frac{4 \times 0.1257}{\pi}} \approx 0.4\, \text{m}
\]
which matches the original pipe diameter, indicating no reduction if the flow velocity remains constant.
Oscillatory Motion and Spring Constant Calculation
A mass of 0.2 kg hung from a spring oscillates with a frequency of 50 Hz when displaced by 0.04 m. The angular frequency (ω) relates to the frequency (f):
\[
\omega = 2 \pi f = 2 \pi \times 50 \approx 314.16\, \text{rad/sec}
\]
Given the formula for simple harmonic motion:
\[
\omega = \sqrt{\frac{k}{m}}
\]
The spring constant:
\[
k = m \times \omega^2 = 0.2 \times (314.16)^2 \approx 0.2 \times 98,696 \approx 19,739\, \text{N/m}
\]
Time for 5 oscillations:
\[
T = \frac{1}{f} = \frac{1}{50} = 0.02\, \text{s}
\]
Total time:
\[
T_{total} = 5 \times T = 5 \times 0.02 = 0.1\, \text{s}
\]
Thus, the mass completes five oscillations in 0.1 seconds.
Trajectory of a Spinning Ball in Flight
The spinning ball's trajectory affected by air resistance and the Magnus effect will curve, resulting in a lateral deviation in the direction of spin. The ball will follow a curved, possibly parabolic path with a deflection toward the direction indicated by the arrow, demonstrating the influence of rotational lift and drag forces acting on the spinning object during flight. To visualize, the trajectory is a curve deviating from the standard parabolic path expected for a projectile without spin, illustrating the complex interplay of aerodynamic forces.
Conclusion
This comprehensive analysis demonstrates foundational principles of physics including buoyancy, fluid mechanics, simple harmonic motion, and projectile dynamics, with calculations illustrating key concepts such as density, flow rate, oscillation period, and trajectory modification due to spin. Understanding these phenomena is essential in engineering design, environmental physics, and sports science, illustrating the valuable intersection of theory and applied physics.
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