An object weighing 300 N in air is immersed in water after being tied to a string connected to a balance. The scale now reads 265 N. Immersed in oil, the object appears to weigh 275 N. Find (a) the density of the object and (b) the density of the oil.
I am unsure of how to proceed with this problem. This is non-calculus based physics so fluidic resistance and mass of the string are negligible. A full solution would be great but the first few steps to how to do it would suffice. And before anyone starts thinking "Oh great another kid who cant do his own work", I am a student taking this physics as an elective. My major is business management and I am doing this because I enjoy it. Thanks a lot for the help!
2007-12-09 10:55:15 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Given the force of gravity ( ~10 N/kg ), you can calculate the mass of the object. 300N => 30kg.
300-265=35 => , the mass of water displaced is 3.5kg. Given the density of water (~1000kg/m^3), you can calculate that the volume displaced, equal to the volume of the object is 3.5 litres.
Density=mass/volume. You should be able to do the rest by yourself.
2007-12-09 11:05:34 · answer #1 · answered by BotsMaster 3 · 0⤊ 0⤋
The difference in what the scale reads is due to the difference in the bouyancy B force. The B force is due to the amount of liquid displaced by the object. Assume that the volume of the object is constant and that the object is completely submerged in the water and in the oil. You know the density of water so you can calculate the density of the object and the oil. Calculate the density of the object first using the info about water.
Good luck and yes this stuff is pretty fun
2007-12-09 11:02:58 · answer #2 · answered by Gary H 7 · 0⤊ 0⤋
a similar mass has 2 different velocities reckoning on the buoyancy distinction, so which you will merely take the adaptation in KE of the two: supply the plastic sphere an arbitrary mass of a million kg KE = a million/2 mv^2 no buoyancy velocity --> KE = 0.5 * a million kg ( 0.36 m/s)^2 KE = 0.0648 J buoyancy velocity--> KE = 0.5 * 1kg (0.24 m/s)^2 KE = 0.0288 J Now merely divide the adaptation in KE in the past buoyancy and after: 0.0288 J / 0.0648 J = 0.44444 So the fraction of the burden it is buoyancy rigidity is (40 4.4% - a hundred% ) = 55.55% of entire rigidity, or 5/9ths
2016-10-01 06:16:03 · answer #3 · answered by ? 4 · 0⤊ 0⤋