Question

A cylindrical piece of metal is 4.5 dm in height with radius of 5.50 x 10^-5 km.

Answer 1

Answer:

a) $V=4.3x10^3mL$

b) $V=4.3x10^6mm^3$

c) $\rho=1.5x10^5g/L$

Explanation:

Hello,

a) In this case, the given height in cm is:

$h=4.5dm*\frac{1m}{10dm}* \frac{100cm}{1m}=45cm$

And the radius in cm is:

$r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{100cm}{1m}=5.5cm$

Thus, the volume in cubic centimeters which is also equal in mL (1cm³=mL) is:

$V=\pi (5.5cm)^2*45cm\\\\V=4.3x10^3cm^3=4.3x10^3mL$

b) In this case, the given height in mm is:

$h=4.5dm*\frac{1m}{10dm}* \frac{1000mm}{1m}=450mm$

And the radius in mm is:

$r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{1000mm}{1m}=55mm$

Thus, the volume in cubic millimeters is:

$V=\pi (55mm)^2*450mm\\\\V=4.3x10^6mm^3$

c) Finally, since 1000 mL equal 1 L, the required density in g/L turns out:

$\rho=\frac{m}{V}=\frac{6.54x10^5g}{4.3x10^3mL}*\frac{1000mL}{1L}\\ \\\rho=1.5x10^5g/L$

Best regards.