Question

pm (3.00, -4.00) m and (700, 7.00) m 14.9

Answer 1

Given : r = (6.30 $\pm$ 0.30) cm , m = (1.71 $\pm$ 0.06) kg

Solution:

Step:1 Determine density

We know that density is given by :

Density = mass / volume

Volume of a sphere , V = (4/3)$\pi$r³

i.e. $\rho$ = m/V

= 1.71 kg / [4/3×3.14×(6.30×10^-2)³]

= 1.63 × 10³ kg/m³

Step:2 Determine uncertainty

Fractional error in the density is given by:

$\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} +\frac{3\Delta r}{r}$

= 0.06/1.71 + 3(0.30)/6.30

= 0.178

Total absolute uncertainty is given by:

(0.178)*(1.63×10³) = 0.29×10³

Step:3 Round and standard form

$\rho$$\pm$$\Delta\rho$ =  (1.63 $\pm$ 0.29)×10³ kg/m³