Question

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)?

Answer 1

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Below is the solution:

37.9*1000*1.03=39037 g=39.037 kg
39.037*2.21=86.28 lb
86.28+59.5=145.78 lb <450 lb

Answer 2

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.