If an object that floats on the surface desplaces 10 cm3 of water how much does that object weigh?

Answer 1

By Archimedes' Principle, a floating body displaces a volume of fluid whose weight is exactly equal to its own.

• This assumes that the body is in equilibrium.
• In order to know the weight of the body (i.e. the object) you must know the density of the fluid in which it is floating.
• Also, Surface Area (SA, in cm2) of object must be known.
• Suppose density of the fluid is rho.

Thus,

Having height = 10cm

Wobject = Volumedisplaced*Densityfluid = SA*Height*rho

Suppose it is water with rho = 1g/cm3, SA = 1cm2

Wobject = 10cm*1cm2*1g/cm3 = 10g.

Just make sure your units are correct.

Answer 2

Since the object is floating (except for the amount that sinks in the water and displaces 10 cc ) the upward bouyant force of the water must equal the object's weight. The upward bouyant force of the water is found from Archimedes' principle: Water will exert an upward bouyant force equal to the weight of the displaced water. Water has a density of 1gram/cc so 10 cc of water equals 10 grams or .01 kg. The weight is mg , which is .01 x 9.8 = .098 Newtons. So the object weighs .098 N which in English units is approximately .4 ounces.

Answer 3

The weight of the object cannot be determined with the given information. The density of the floating object must be determined in order to solve for its weight. According to Archimedes' principle, the weight of an object in a fluid is equal to the quotient of the density of the object and the density of the fluid multiplied by the weight of the fluid displaced.

Assuming that the density of water is equal to 1g/cm3, then,

(The Weight of the Object) = (The Density of the Object) x 10

Answer 4

10gms