Answer :
Answer: 450 lb > 145.38 lb ,Yes. Metal stand will able to support the aquarium with 37.9 L of water.
Explanation:
Maximum load sustained by metal stand = 450 lb
Weight of the aquarium = 59.5 lb
Density of seawater = 1.03 g/mL
Volume of the seawater = 37.9 L = 37900 ml (1 L=1000 mL)
[tex]Density=\frac{\text{mass of water}}{\text{volume of the water}}=1.03 g/mL=\frac{mass}{37900 mL}[/tex]
Mass of the water = 39,037 g = 85.88 lb (1 g = 0.0022 lb)
Load due to aquarium and water on metal stand :
= 59.5 lb + 85.88 lb = 145.38 lb
450 lb > 145.38 lb Yes. Metal stand will able to support the aquarium with 37.9 L of water.
Final answer:
After calculating the total weight of the aquarium and the seawater (145.57 lb), it is clear that the metal stand can support it since it is less than the stand's maximum load capacity of 450 lb.
Explanation:
The student asks if a metal stand that can support a maximum load of 450 lb will be able to hold an aquarium weighing 59.5 lb and filled with 37.9 liters of seawater (density = 1.03 g/ml). To solve this, we need to calculate the total weight of the aquarium when filled with seawater and compare it with the stand's maximum load capacity.
First, let's convert seawater volume to weight:
37.9 L = 37,900 mL (since 1 L = 1,000 mL)
The total weight of seawater (in grams) = 37,900 mL * 1.03 g/mL = 39,037 g
Convert seawater weight to pounds: 39,037 g * (1 lb / 453.592 g) ≈ 86.07 lb
Next, add the weight of the aquarium to the weight of the seawater:
Total weight = Aquarium weight + Seawater weight
Total weight = 59.5 lb + 86.07 lb ≈ 145.57 lb
Since 145.57 lb is less than the maximum load of 450 lb, the metal stand will support the aquarium with the seawater.
