I need to know if the weight of a pebble would remain constant in air and in water. Thanks!
2007-07-18 06:41:13 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Physics
scales weigh "apparent weight" not actual weight.
apparent weight equals the sum of forces acting on an object including bouyancy, gravity, centrifugal forces, magnetic, etc. when you place a rock in water, there is an upward force equal to the weight of water displaced acting on the rock. so the apparent weight would change by the weight of the water displaced (it would be less).
actual weight, by definition, is the force due to gravity on an object. F = mg, where F = force (or weight), m = mass, g = acceleration due to gravity. in effect, the actual weight ignores the rest of the forces acting on the rock.
so, the answer is...
apparent weight would be not be constant
actual weight would be constant.
2007-07-18 07:00:43 · answer #1 · answered by Dr W 7 · 1⤊ 0⤋
Obviously the mass of the pebble is still the same but the weight of the pebble would be offset by the mass of the water pushed aside by the stone.
This is the force that makes it possible for ships to stay afloat. A ship will float as long as the mass of the water pushed aside by the hull is greater than the mass of the ship itself.
The difference in weight of your pebble is equal to the volume of the pebble times the mass of water at any given temperature.
Hope this helps.
2007-07-18 13:51:09 · answer #2 · answered by Ronny K 1 · 0⤊ 2⤋
It would weight less in water ... obviously ... think of trying to weight ice ... you could not even do it because it floats in water. Water or any liquid, and even the air provides a bouyancy effect, removing it's equivalent weight(mass) of its density * the volume of whatever you are weighing it in.
If you figure out the density of air, and the volume of your body you realize why we do not care about that tiny difference ... but in water, it is significant, especially if what you are tryng to weight is less dense than water.
2007-07-18 13:48:44 · answer #3 · answered by themountainviewguy 4 · 1⤊ 1⤋
No, the weight of rock in air and in water would be different.
In water, the observed weight of rock would be less than that in air & this is because of the upward thrust force exerted by the water on the rock.
Any body when immersed in water experiences an upward thrust due to which it experiences weighlessness.
2007-07-18 13:48:07 · answer #4 · answered by deepak_hellboy 1 · 1⤊ 2⤋
no. when we put one thing in the water or any liquid the liquid will press it to up,so this force will reduce the weight of rock,(because this force is opposite of weight force)but in the air this pressure is alittle and the weight of rock is more than in the water.
2007-07-18 14:11:01 · answer #5 · answered by Anonymous · 0⤊ 1⤋
No, it would weigh less in water, because it displaces water (moves it away) and the water acts to "buoy" it up (even though it sinks anyway). The displaced water acts to force the pebble up, so it weighs less.
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2007-07-18 13:46:48 · answer #6 · answered by tlbs101 7 · 1⤊ 2⤋
No. It's weight in water will be lesser.
Reason: Principle of Buoyancy, and upward thrust.
The explaination given above is abosultely baseless.(Some time is history, people believed in this theory, but was soon discarded as a failure theory)
2007-07-18 13:57:13 · answer #7 · answered by MDA 4 · 0⤊ 1⤋
No, it would not be the same due to the difference in the density of water and air.
2007-07-18 13:46:12 · answer #8 · answered by zeth74 1 · 1⤊ 2⤋
No, it will weigh more in water, because of the pressure of the water. The same as with the pull of gravity(oxygen) on the surface of the earth (which will cause it to weigh more on the surface than in water).
2007-07-18 13:54:54 · answer #9 · answered by bigmama35 3 · 0⤊ 4⤋
no - it would weigh less.. the water puts a 'buoyant force' on the object causing it to weigh less in water..
2007-07-18 13:49:40 · answer #10 · answered by miggitymaggz 5 · 1⤊ 1⤋