the radius of a solid sphere is measured to be (6.50+-0.20)cm,and its mass is measured to be (1.85+-0.02)kg.determine the density of the sphere in kilogram per cubic meter and its uncertainty.不要剪接.万份感激.不要简捷* ；用英文解答

the radius of a solid sphere is measured to be (6.50+-0.20)cm,and its mass is measured to be (1.85+-0.02)kg.determine the density of the sphere in kilogram per cubic meter and its uncertainty.不要剪接.万份感激.不要简捷* ；用英文解答
the radius of a solid sphere is measured to be (6.50+-0.20)cm,and its mass is measured to be (1.85+-0.02)kg.determine the density of the sphere in kilogram per cubic meter and its uncertainty.不要剪接.万份感激.
不要简捷* ；用英文解答为佳〉〉〉〉 为何δV=3δr=0.093？

the radius of a solid sphere is measured to be (6.50+-0.20)cm,and its mass is measured to be (1.85+-0.02)kg.determine the density of the sphere in kilogram per cubic meter and its uncertainty.不要剪接.万份感激.不要简捷* ；用英文解答
半径r=6.50±0.20 cm
质量m=1.85±0.02 kg
相对误差δr=0.20/6.50=0.031
δm=0.02/1.85=0.011
V=4πr^3/3=4*3.14*6.5^3/3=1150 cm^3
δV=3δr=0.093
ρ=m/V=1.85/1150*10^-6=1.61*10^3 kg/m^3
δρ=δm+δV=0.104
ρ*δρ=1.61*10^3*0.104=0.17*10^3 kg/m^3
所以：该物体密度为(1.61±0.17)*10^3 kg/m^3.

可以用英文解答不？？？？？

我帮你译下，剩下的自己想吧，几年没做，都忘记得差不多了：
球体半径为 (6.50+-0.20)cm，质量为(1.85+-0.02)kg。求kg/m3的密度和误差。

我汗死~

Okay, I am not really good at explain this, but, normally, you need to know the volume of a sphere:
Volume = (4/3) (pi)(radius)^3
so, substitude the numbers in:
V= 4/3 X Pie X (6.50/100)c...

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Okay, I am not really good at explain this, but, normally, you need to know the volume of a sphere:
Volume = (4/3) (pi)(radius)^3
so, substitude the numbers in:
V= 4/3 X Pie X (6.50/100)cube which equals 1.1503...X10^-3
You then can calculate the density of the sphere by using the formulae: Density = mass / vol
so, 1.85kg divided 1.1503...X10^-3 by which equals 1608.21107...
Now, we can calculate the uncertainty in the density.
Absolute uncertainty of the density is given by:
Δρ = √[ (∂ρ/∂m)²∙(Δm)² + (∂ρ/∂r)² ∙ (ΔR)² ]
= √[ (m/((4/3)∙π∙R³))²∙(Δm)² + (-3∙m/((4/3)∙π∙R⁴)
)² ∙ (ΔR)² ]
It's easier to calculate the relative uncertainty:
Δρ/ρ = √[ (1/((4/3)∙π∙R³))²∙(Δm)² + (-3∙m/((4/3)∙π∙R⁴))² ∙ (ΔR)² ] / m/((4/3)∙π∙R³)
= √[ (Δm/m)² + (-3)²∙ (ΔR/R)² ]
= √[ (0.02/1.85)² + 9 ∙ (0.2/6.5)² ]
= 0.05842677...
=5.84%
Hence:
ρ = 0.0584266.. ∙ 1608.211.../m³ = (about a hundred and 10 ish)kg/m³
(please check the calculation of each one, i just did it without a calculator so it's not really accurate, but i am pretty sure that the methods are correct.)
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