1) When a 61 N stone is attached to a spring scale and is submerged in water, th
ID: 1484053 • Letter: 1
Question
1)
When a 61 N stone is attached to a spring scale and is submerged in water, the spring scale reads 40 N. What is the density of the stone?
kg/m3
1)
A homogeneous solid object floats on water with 82% of its volume below the surface. The same object when placed in a second liquid floats on that liquid with 72% of its volume below the surface.
Determine the density of the object.
kg/m3
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2)
Determine the specific gravity of the liquid.
Explanation / Answer
1) solution: The buoyant force, Fb, equals to the weight, W, of water displaced by the stone..
And is therefore equal to the difference between readings of the spring scale and
the actual known weight of the stone. (61 – 40) = 21 N
The weight of the displaced water is given by::
W = mg = Vg
where
volume of displaced water: V, same as the volume of the stone.
density of water: = 1000 kg/m³
V = W / ( g)
V = 21 / 1000*9.81) = 2.140x10-³ m³ = 2.140 liters
The density of the object is:
= W/(gV) = 61/(9.81*2.140x0.1³)
= 2905.67 kg/m³
2)assume it has a volume of 1 m³. Then the part below the surface is 0.82 m³.
0.82 m³ of water weighs 820 kg. That means the weight of the object is 820 kg, since it is balanced by the upwards pressure of the displaced water.
Therefor the density is weight/volume = 820 kg/m³
b) Specific gravity is a ratio of densities with the reference desity being water: 1000kg/m^3
We know the density of the block. We can calculate the density of the unknown liquid since we know that a volume of the unknown liquid = to 72% of the blocks volume weighs what the block does. We can write: Vx*820*g = 0.72*Vx*Dl*g where Dl is the density of the unknown liquid. Vx and g both appear on both sides of the equation so they drop out leaving
820=0.72*Dl => Dl = 820/0.72 = 1138.89 kg/m^3 which means it has higher density than water because the block floats even higher in this liquid than it does in water. So this density makes sense
The specific gravity is the ratio of the unknown liquid to that of water:
1138.89/1000 = 1.389
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