Question
- mylabmastering.pearson.com /?courseld=13243292#/ A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.80 cm below the interface (Figure 1). The density of the oil is 790kg/m^3 For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Buoyancy. p=p_(interf) p-p_(a)= p-p_(a)= p-p_(a)=9 Part C What is the mass o Express your ans
Solution
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Emersyn
Veteran · Tutor for 10 years
Answer
Here's how to solve this buoyancy problem:**1. Forces Acting on the Block:**The block is in equilibrium, meaning the forces acting on it are balanced. These forces are:* **Weight (W):** Downward force due to gravity.* **Buoyant Force from Oil (B_oil):** Upward force due to the displaced oil.* **Buoyant Force from Water (B_water):** Upward force due to the displaced water.**2. Expressing the Forces:*** **Weight (W):** W = m * g = ρ_wood * V * g (where m is the mass of the wood, ρ_wood is the density of the wood, V is the volume of the wood, and g is the acceleration due to gravity).* **Buoyant Force from Oil (B_oil):** B_oil = ρ_oil * V_oil * g (where ρ_oil is the density of the oil, and V_oil is the volume of oil displaced).* **Buoyant Force from Water (B_water):** B_water = ρ_water * V_water * g (where ρ_water is the density of water (1000 kg/m³), and V_water is the volume of water displaced).**3. Calculating the Volumes:*** **Total Volume (V):** V = (0.10 m)³ = 0.001 m³ (convert cm to m)* **Volume of oil displaced (V_oil):** V_oil = (0.10 m)² * (0.082 m) = 0.00082 m³ (The block is 10cm high, and 8.2cm is submerged in oil)* **Volume of water displaced (V_water):** V_water = (0.10 m)² * (0.018 m) = 0.00018 m³**4. Setting up the Equilibrium Equation:**Since the block is floating, the upward buoyant forces equal the downward weight:W = B_oil + B_waterρ_wood * V * g = ρ_oil * V_oil * g + ρ_water * V_water * g**5. Solving for the mass of the wood (m):**We know W = m * g, so we can rewrite the equation and solve for m:m = (ρ_oil * V_oil + ρ_water * V_water)m = (790 kg/m³ * 0.00082 m³ + 1000 kg/m³ * 0.00018 m³)m = 0.6478 kg + 0.18 kgm = 0.8278 kg**Answer:**The mass of the wood block is approximately 0.828 kg. You should round to the appropriate number of significant figures as required by your instructor.