First calculate the corresponding mass of 37.9 L
M = 37.9 L ( 1000 ml / 1 L ) ( 1.03 g / ml)
M = 39,037 g
Convert g to pounds
M = 39,037 g ( 1 lb / 453.592 g)
M = 86.06 lb
Then add the 59.5 ln
M = 86.06 + 59.5 = 145.56 lb so it will support it
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.
