Respuesta :
Answer:
Explanation:
Given
weight of anchor [tex]W=250\ pounds[/tex]
weight density [tex]\rho =480\ lbs/ft^3[/tex]
volume of anchor [tex]V=\frac{250}{480}=0.520\ ft^3[/tex]
density of sea water [tex]\rho _s=64\ lbs/ft^3[/tex]
buoyant force on anchor when immersed in sea water [tex]=\rho _s\times V=33.28 lbs[/tex]
Force required to lift the anchor=Weight-Buoyancy
[tex]F=250-33.28=216.72\ lbs[/tex]
Answer:
Explanation:
Weight in air = true weight = 250 pounds
density of iron , d = 480 lbs/ft³
density of sea water, d' = 64 lbs/ft³
Volume of iron = mass / density of iron
V = 250 / 480 = 0.521 ft³
Force required to lift = weight of iron in water = Apparent weight of iron
Weight in water = true weight - buoyant force
Weight in water = 250 x g - volume x density of water x g
Weight in water = 250 g - 0.521 x 64 x g
Weight in water = 216.67 g
Weight in water = 216.67 pounds
Thus, the force required to lift the iron is 216.67 pounds.
