Yes, assuming that immersed object has no internal voids which the fluid cannot fill (e.g. a hollow sphere).
Archimedes's principle states that the buoyant force acting on an object immersed or floating in a fluid equals the weight of the fluid displaced.
Archimedes' Principle is "The buoyant force acting on an object equals the weight, (force of gravity) of the fluid displaced by the object. (Answer found in sciencepowere grade 8 textbook.)
What you are describing is Archimedes' principal. The reason it holds try is that the object will only float when the force of gravity pulling it down equals the force due to buoyancy. This happens once the body has displaced its mass worth of the fluid.
The conclusion of the Archimedes principle is simply that the upward buoyant force that is experienced by a body immersed in a fluid, is equivalent to the weight of the fluid that the body displaces. This allows the volume of an object to be measured by measuring the volume of liquid it displaces after submerging. For any immersed object, the volume of the submerged portions equals the volume of fluid it displaces.
Archimedes Principle simply states that the buoyant force acting on an object equals the weight (force of gravity) of the fluid displaced by the object. He came to formulate it when he was in the tub, he noticed that before he got in, the water was at the rim and after he got in, the water had spilled over the top.
Not at all. (The buoyancy force equals the weight if the displaced water,)
It's equals to weight of fluid it displaces
Well, as an object is in a fluid, it displaces the water and more of one or more different objects same or different will cause more water displacement.
Buoyancy.The Archimedes' Principle establishes that "any object inside a fluid experiments a force equals to the weight of the liquid displaced by the body" or, mathematically:F = (rho)·g·Vwhere rho is the density of the fluid, g the gravity acceleration and V the volume displacement exerted by the body.
Weight vs volume. PLace object in container already full of water. Measure volume of liquid displaced. weight object. weight divided by volume equals density
The volume of the displaced water would be less - as you're reducing the mass of the boat. Another viewpoint: I think there's a bit more to this question, but the basic answer remains the same. I think it's all about "Archimedes' Principle". Let's consider the anchor as still part of the boat. Also let's think about it before it gets partly buried in the ground underwater. Archimedes' Principle tells us: The "upthrust" on the boat before the anchor is lowered is equal to the weight of the whole boat. That equals the weight of water displaced. The anchor itself doesn't float in water. It is denser than water. When the anchor is completely submerged it displaces an amount of water equal to the anchor's volume (not the anchor's weight). When the anchor was on the boat it displaced an amount of water equal to its weight. So, when the anchor is lowered, the boat (including anchor) displaces a slightly smaller volume of water.
A floating ship displaces its weight in water and the water pushes back (according to newton's 3rd action-reaction law). The upward force is called upthrust. This occurs according to Archimedes' Principle which states that when a body is full or partially submerged in a fluid (water), the upthrust equals the weight of the water displaced.