Finding the Density of a given object

  • Thread starter Thread starter iJpawn
  • Start date Start date
  • Tags Tags
    Density Volume
AI Thread Summary
The density of an object is calculated by dividing its mass by its volume. The mass of Earth is approximately 6x10e+24 kg, and its radius is 4000 miles, which converts to about 6400 km. The volume is calculated using the formula V=4/3π(r)^3, leading to a volume of approximately 1x10e+12 m^3 after proper unit conversion. The resulting density calculation of 6x10e^12 kg/m^3 is incorrect, as it is too high, indicating a possible error in the volume conversion from cubic kilometers to cubic meters. It is recommended to show all calculations with units clearly to identify any mistakes.
iJpawn
Messages
3
Reaction score
0
The density of an object equals its mass divided by its volume. The mass of Earth is 6x10e+24 and its radius is 4000 miles (1.61km = 1mile). What is the density of the Earth in kg/m^3?V=4/3pi(r)^3. D=M/VConverting 4000 to Km, I get 6400. After plugging it into the Volume equation, I end up with approximately 1x10e+12. I divide 6x10e+24 by 1x10e+12, and my result is 6x10e^12. Is this the correct way to do the given problem? My answer sheet from my professor does not state this as any of the answers, and I can't think of any other way to approach this equation.
 
Physics news on Phys.org
The answer should be in kg/m^3.
 
iJpawn said:
The density of an object equals its mass divided by its volume. The mass of Earth is 6x10e+24 and its radius is 4000 miles (1.61km = 1mile). What is the density of the Earth in kg/m^3?V=4/3pi(r)^3. D=M/VConverting 4000 to Km, I get 6400. After plugging it into the Volume equation, I end up with approximately 1x10e+12. I divide 6x10e+24 by 1x10e+12, and my result is 6x10e^12. Is this the correct way to do the given problem? My answer sheet from my professor does not state this as any of the answers, and I can't think of any other way to approach this equation.
Your calculation is hard to follw if you do not state the units at every point. The answer is much too high.
I suspect you erred in converting cu km to cu m.
What units are the given mass in?
 
The mass of Earth is given in 6x10e+24kg, sorry for not posting it earlier!
 
iJpawn said:
The mass of Earth is given in 6x10e+24kg, sorry for not posting it earlier!
Ok, but repost your calculation showing units at every step. Omit no steps.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top