Because any object in water is buoyed up by a force equal to
the weight of the water it displaces (pushes aside).
The weight of a rock is less in water because water exerts an upward buoyant force on the rock, reducing its effective weight. This is due to the principle of buoyancy, where the weight of the water displaced by the rock counteracts the rock's weight.
The weight of the rock will be lessened by the weight of the same volume of water. If the rock weighs less than that volume of water, it will float (as does pumice). Thus, if you had a rock that has a density (weight/volume) of 2 times the same volume of water, it would weigh one half of its dry weight when in water.
No, a rock does not weigh more in water. When submerged in water, the rock displaces an amount of water equal to its own volume, leading to a buoyant force that reduces its apparent weight. So, the rock will weigh less in water compared to in air.
it is displacing more water imagine putting a rock into a small container of water. the container appears to have more water because it was displaced
Since the weight of pumice is less it will float on water for some time and then..
It depends what the weight reading was originally measuring. If it was measuring the weight of the experimenter and the rock they were holding, and the water is not being held by them, then the weight will decrease by the weight of the rock. If it was measuring the weight of the water into which the rock it dropped, then it will increase by the weight of the rock. If it was measuring the weight of something totally unrelated to the experiment, then dropping the rock will have no measurable effect on the reading of the weight. Context needs to be given for the weight reading for a proper answer to be given.
The buoyant force acting on a rock in water is equal to the weight of the water displaced by the rock, as described by Archimedes' principle. This means that the buoyant force is equal to the weight of the volume of water that is pushed aside by the rock when it is submerged.
The buoyant force acting on the rock submerged in water is equal to the weight of the water displaced by the rock. This is known as Archimedes' principle, which states that the buoyant force on an object is equal to the weight of the fluid it displaces.
A rock floats in water due to buoyancy, which is the upward force exerted by the water on the rock. If the rock is less dense than the water, it will displace an amount of water equal to its weight, causing it to float. Additionally, surface tension can also play a role in helping small rocks or pebbles float on the surface of water.
The buoyant force accounts for the missing 2 N when the rock is in water. The 2 N is the weight of the volume of water equal to the volume of the rock ... the water that the rock 'displaces' (pushes aside) when it enters the water.
The buoyant force of water will partially counteract the weight of the rock, making it feel lighter. The force required to move the rock will be less than 50 pounds, as a portion of its weight is offset by the buoyant force. The exact force needed will depend on the shape and density of the rock, and the friction between the rock and the surface it is resting on.
A rock will sink because it is more dense than the liquid into which it is placed.