AAMC 4 Discrete # 18
An object with 15 grams mass is immersed in benzene and suffers an apparent loss of mass of 5 grams. What is the approximate specific gravity of the object? (Sp. gravity of benzene = 0.7)
A. 1.4
B. 1.8
C. 2.1
D. 3.0
I lucked out with this. I'd really like if somebody goes over my work because my answer has the wrong number of 0's!
My work:
15 g = 0.15 N
5g = 0.05 N
0.05 N pushing is the buoyant force.
Fb = 0.005 = V*rho(benzene)*g = V * 700*10
V = 0.005 / 7000
rounding to hell, I put the 10's aside and work with the important numbers:
5/7 *10^-6 ~ a bit more than half, say, 0.7 * 10^-6 = 6 * 10^-7 meters cubed
Density = mass / volume = 0.015 / 7*10^-7 = 1.5 / 7 * 10^6 = 2.1* 10^6
*10^6!!! it should be *10^3!
An object with 15 grams mass is immersed in benzene and suffers an apparent loss of mass of 5 grams. What is the approximate specific gravity of the object? (Sp. gravity of benzene = 0.7)
A. 1.4
B. 1.8
C. 2.1
D. 3.0
I lucked out with this. I'd really like if somebody goes over my work because my answer has the wrong number of 0's!
My work:
15 g = 0.15 N
5g = 0.05 N
0.05 N pushing is the buoyant force.
Fb = 0.005 = V*rho(benzene)*g = V * 700*10
V = 0.005 / 7000
rounding to hell, I put the 10's aside and work with the important numbers:
5/7 *10^-6 ~ a bit more than half, say, 0.7 * 10^-6 = 6 * 10^-7 meters cubed
Density = mass / volume = 0.015 / 7*10^-7 = 1.5 / 7 * 10^6 = 2.1* 10^6
*10^6!!! it should be *10^3!