Submerged Block

Related BrainMass Solutions

• Buoyant Force in Equilibrium of Submerged Block

  20535 Buoyant Force in Equilibrium of Submerged Block A block of plastic whose mass is 25 kg and whose density is 850 kg/m^3 is tied by a corfd to the bottom of a pool of seawater whose density is 1025 kg/m^3. Find the force in the cord.

• Buoyant force as applied to an iron block submerged in water

  Other one is the weight of the block Buoyant force is the weight of the liquid displaced by the submerged part. Here the whole block is submerged.

• Archimedes principle to find density of a block floating in water and density of the oil

  17592 Finding the Density of a Block and Oil A block of wood floats in fresh water with forty percent of its volume submerged. This block floats in oil with 93 percent of its volume submerged. (a) Find Db, the density of the block.

• Archimedes principle

  same amount of wood is submerged in water as before (b) If 75% of the wood is submerged in water before the oil is added, find the fraction submerged when oil with a density of 731 kg/m3 covers the block.

• Buoyant Force of Block

  32206 Buoyant Force of Block A 227-kg block of cement, of density 2.8 x 10^3 kg/m^3, rests on a pedestal in front of the Al Capone Memorial Library. How much did it weigh submerged, while being hauled out of the river (freshwater)?

• Archimedes' principle applied to a floating block in turpentine gives density of liquid.

  6606 Archimedes' principle applied to a floating block in turpentine gives density of liquid. Archimedes' principle applied to a floating block in turpentine gives density of liquid. The same block, held stationary submerged by a metal block.

• Archimedes' principle and equilibrium of two submerged objects to find density

  It is tied by a cord to a block of aluminum, mass Ma= 1.5 kg, whose density is Da= 2700 kg/m^3. The two masses are submerged in sea water, density Dliq=1025 kg/m^3, and held stationary by a cord exerting upward force C= 6.57 nt.

• An object is held stationary by a cable near the sea floor. Find the cable tension.

  .= DL g Vsub, in which DL is the liquid density, Vsub is the submerged volume, and g is 9.8 nt/kg, and using the relationship 'density = mass per volume', D= m/V, we can write: DL g (M/DB) = M g + C.

• Volume and density problem

  28263 Volume and density problem A block of wood half as dense as water floats with half its volume above water. A piece of iron is then tied on top so the wood floats with only 1/4 its volume above surface.

• Evaluating weight from mass of displaced water

  The volume of water is the volume of the submerged part of the block: Volume of water = 0.6 * 0.4 m^3 = 0.24 m^3 Density of water = 1000Kg/ m^3 So the weight of ater dispalced = 0.24 * 1000 = 240 Kg This is the weight of the mass..