Asked by Some kid
For a liquid, 0.500 cubic meters weighs 4590 N. What is the specific gravity of the liquid?
Answers
Answered by
DrBob222
density = mass/volume
d = 4590/0.500 = ?? N/m^3
d = 4590/0.500 = ?? N/m^3
Answered by
GK
Here is a quick rundown of the way I think it should be done:
If the weight is 4590 N,
m = W/g, or
m = 4590 N / 9.8 kg/N = 468.4 kg
d = 468.4 kg / 0.500 m^3 = 937 kg/m^3
1 m^3 = 1000 dm^3
0.500 m^3 = 500 dm^3
d = 468.4 kg / 500 dm^3 = 0.937 kg/dm^3
It can be shown that a density of
0.937 kg/dm^3 is equivalent to 0.937 g/cm^3
The specific gravity is the ratio of the densities of the material divided by the density of water:
0.937 g/cm^3 / 1.00 g/cm^3 = 0.937 (unitless value)
If the weight is 4590 N,
m = W/g, or
m = 4590 N / 9.8 kg/N = 468.4 kg
d = 468.4 kg / 0.500 m^3 = 937 kg/m^3
1 m^3 = 1000 dm^3
0.500 m^3 = 500 dm^3
d = 468.4 kg / 500 dm^3 = 0.937 kg/dm^3
It can be shown that a density of
0.937 kg/dm^3 is equivalent to 0.937 g/cm^3
The specific gravity is the ratio of the densities of the material divided by the density of water:
0.937 g/cm^3 / 1.00 g/cm^3 = 0.937 (unitless value)
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