Answer:
this is how u solve this
Step-by-step explanation:
Answer:
The weight of the water in the pool is approximately 60,000 lb·f
Step-by-step explanation:
The details of the swimming pool are;
The dimensions of the rectangular cross-section of the swimming pool = 10 feet × 20 feet
The depth of the pool = 5 feet
The density of the water in the pool = 60 pounds per cubic foot
From the question, we have;
The weight of the water in Pound force = W = The volume of water in the pool given in ft.³ × The density of water in the pool given in lb/ft.³ × Acceleration due to gravity, g
The volume of water in the pool = Cross-sectional area × Depth
∴ The volume of water in the pool = 10 ft. × 20 ft. × 5 ft. = 1,000 ft.³
Acceleration due to gravity, g ≈ 32.09 ft./s²
∴ W = 1,000 ft.³ × 60 lb/ft.³ × 32.09 ft./s² = 266,196.089 N
266,196.089 N ≈ 60,000 lb·f
The weight of the water in the pool ≈ 60,000 lb·f
<span>Surface Area =<span> (2 • <span>π <span>• r²) + (2 • <span>π • r • height)
</span></span></span></span></span>
<span>(2 • <span>π <span>• r²) is surface area of the "ends"
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<span>(2 • <span>π <span>• r • height) is lateral area
</span></span></span>
<span>Lateral Area = 2 * PI * 2 * 7
</span>
<span>Lateral Area =28 * PI =
</span>
<span><span><span>87.9645943005 square inches</span></span></span><span><span><span>
</span>
</span>
</span>
Answer is B
Answer:
What do u mean
Step-by-step explanation:
