Question

What will be the density (p) of iron cube of mass (m) of metal =5kg volume (v)=685 cm cube answer?

Answer 1

Answer:

Step-by-step explanation:

D=m/v

m=5 kg

V=685 cm cube =0.000685 m cube

D = 5/0.000685

D=7299

Answer 2

The density of iron ($\rho$) = $7.299 \times 10^{-3} \dfrac{kg}{m^3}$

Step-by-step explanation:

Given,

The mass of iron (m) = 5 kg and

The voume of iron (m) = 685 $cm^{3}$ = 685 × $10^{-6} m^3$

To find, the density of iron ($\rho$) = ?

We know that,

The density of iron ($\rho$) = $\dfrac{Mass}{Volume}$

= $\dfrac{5}{685\times 10^{-6}} \dfrac{kg}{m^3}$

= $\dfrac{5}{685\times 10^{-6}} \dfrac{kg}{m^3}$

= $0.007299 \times 10^{-6} \dfrac{kg}{m^3}$

= $7.299 \times 10^{-3} \dfrac{kg}{m^3}$

Thus, the density of iron ($\rho$) = $7.299 \times 10^{-3} \dfrac{kg}{m^3}$