Answer:
14.1 kg
Explanation:
Given:
Length=7.00inches
Width=3.63 inches
Height=1.75 inches
density = 19,300 kg/m3.
We can convert the given parameters to metre for unit consistency
But we know 1 inches= 0.0254 metre
✓Then Length l=7.00inches
=7×0.0254 metre=0.1778m
✓Width w =3.63 inches
==3.63 ×0.0254 metre=0.092m
✓Height h =1.75 inches
=1.75 ×0.0254 metre=0.0445 m
But Mass= density × volume
Volume= Length× width×height
Mass= density× Length× width×height
= 19300kg/m³×0.1778×0.0922×0.0445
=14.1 kg
Therefore, the mass of the gold bar is 14.1 kg
The mass of such a gold bar is of 13.89 kg.
Given data:
The length of gold bar is, [tex]L=7.00 \;\rm in =7.00 \times 0.0254=0.1778 \;\rm m[/tex].
The width of gold bar is, [tex]w= 3.63 \;\rm in =3.63 \times 0.0254 = 0.092 \;\rm m[/tex].
The height if gold bar is, [tex]h = 1.75 \;\rm in =0.044 \;\rm m[/tex].
The density of the gold bar is, [tex]\rho =19,300 \;\rm kg/m^{3}[/tex].
The given problem is based on the concept of density. The density of any substance is equal to the ratio of mass and volume. Considering the gold bar to rectangular shape, the volume of gold bar is calculated as,
[tex]V= L \times w \times h\\\\V = 0.1778 \times 0.092 \times 0.044\\\\V=7.197 \times 10^{-4} \;\rm m^{3}[/tex]
Now, use the formula of density to calculate the mass of gold bar as,
[tex]\rho =\dfrac{m}{V}\\\\m = \rho \times V\\\\m = 19300 \times (7.197 \times 10^{-4})\\\\m= 13.89 \;\rm kg[/tex]
Thus, we can conclude that the mass of such a gold bar is of 13.89 kg.
Learn more about the density of substance here:
https://brainly.com/question/8760548
