To verify her suspicion that a rock specimen is hollow,a geologist weighs the specimen in air and in water.She finds that the specimen weighs twice as much in air as it does in water.The solid part of the specimen has a density of 5 multiplied by 10cubedkg/meter cubed.What fraction of the specimen's apparent volume is solid
Lets make it simple: Let the size of the rock in air have a mass of 200kg, in water, half that.
So the bouyancy from water is 100kg*g, meaning the rock volume is 100kg/densitywater= 100kg/(1000kg/m^3)=.1m^3
now, the rock consist of two parts: hollow, and rock.
volume rockpart=massrock/density=100kg/(5E3kg/m^3= 20E-3 m^3
fraction which is solid= .020/.1=20 percent. Check my work
fraction which is solid=
A mass m = 4.3 kg is attached to a vertical spring with k = 210 N/m and is set into motion. What is the frequency of the oscillation?
Volume of solid in sample is 40%
Density of sample = 2 (from informaton)
If,for example, sample is 100cm3:
Volume(solid)=40 cm3 (from above)
Since density of rock is 5, therefore mass of rock (from 40 cm3)=200g
Therefore density of sample=200/100=2
This verifies the above density for the sample (according to info provided)
To find the fraction of the specimen's apparent volume that is solid, we first need to determine the density of the specimen.
Let's denote the density of the specimen as ρ_specimen and the density of water as ρ_water.
In this case, we are given that the solid part of the specimen has a density of 5 × 10^3 kg/m^3 and the specimen weighs twice as much in air as it does in water. This information allows us to set up the following relationship:
ρ_specimen * V_specimen = 2 * ρ_water * V_specimen
Where V_specimen represents the volume of the specimen.
Canceling out V_specimen from both sides of the equation, we get:
ρ_specimen = 2 * ρ_water
Since the density of water is approximately 1000 kg/m^3, we can substitute this value into the equation:
5 × 10^3 kg/m^3 = 2 * 1000 kg/m^3
Simplifying the equation, we find:
5 × 10^3 = 2 × 10^3
Now, let's divide both sides of the equation by 10^3:
5 = 2
Since the equation is not balanced, there seems to be a mistake in the given information about the density of the solid part of the specimen. It is not possible for the density of the solid to be higher than the density of water, as the specimen would then sink.
Please double-check the provided information or provide additional details if available to continue solving the problem.