Answer:
Step-by-step explanation:
Givens
First, we find the volume of the figure.
[tex]V=\frac{1}{3} (B \times h)[/tex]
Where [tex]B=80cm^{2}[/tex] and [tex]h=7.5 cm[/tex].
Replacing values, we have
[tex]V=\frac{1}{3}(80)(7.5)=200cm^{3}[/tex]
The density is defined as
[tex]\rho=\frac{m}{V}[/tex]
Where we need to find the mass [tex]m[/tex], we already know that [tex]\rho=1.7 \ gr/cm^{3}[/tex] and [tex]V=200cm^{3}[/tex]. Replacing these values, and solving for [tex]m[/tex]
[tex]1.7=\frac{m}{200}\\ m=1.7(200)=340 \ gr[/tex]
Then, we use the weight definition, which is
[tex]W=mg[/tex]
Where [tex]m=340 \ gr=0.34 \ kg[/tex] and [tex]g=9.81 m/sec^{2}[/tex]. Replacing these values, we have
[tex]W=0.34(9.81)\\W=3.33 N[/tex]
Therefore, the weight of Keenan's paperweight is 3.33 N.
