A metal stand specifies a maximum load of 450 lb. Will it support an aquarium that weighs 59.5 lb and is filled with 37.9 L of seawater (d = 1.03 g/mL)?
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Below is the solution:
37.910001.03=39037 g=39.037 kg
39.037*2.21=86.28 lb
86.28+59.5=145.78 lb <450 lb
Answer:
Yes, metal stand will easily support the aquarium filled with 37.9 liters of seawater.
Explanation:
Maximum load taken by the metal stand = 450 lb
Load exerted by the aquarium filled with sea water :
Mass of an empty aquarium = 59.5 lb
Mass of sea water = m
Volume of sea water ,v= 37.9 L = 37900 mL
Density of the sea water = d = 1.03 g/mL
\(Density=\frac{Mass}{Volume}\)
\(Mass=density\times Volume\)
\(m=1.03 g/mL\times 37900 L=39,037 g = 86.062 lb\)
1 lb = 453.592 g
Total load exerted by filled aquarium = 86.062 lb + 59.5 lb = 145 .56 lb
450 lb > 145 .56 lb
Yes, metal stand will easily support the aquarium filled with 37.9 liters of seawater.