Note: the text says that the density of the block is [tex]9000 kg/m^3[/tex] (not [tex]9000 kg/cm^3[/tex], which not a plausible value)
Answer:
63.9 N
Explanation:
We want to find the apparent weight of the block when it is in water.
First of all, we know its true weight:
W = 70 N
So we can find the mass of the block:
[tex]m=\frac{W}{g}=\frac{70}{10}=7.0 kg[/tex]
where [tex]g=10 m/s^2[/tex] is the acceleration of gravity.
From the mass and the density, which is
[tex]\rho=9000 kg/m^3[/tex]
we find the volume of the block:
[tex]V=\frac{m}{\rho}=\frac{7.0}{9000}=7.8\cdot 10^{-4} m^3[/tex]
We know that when the block is immersed in paraffin, it is acted upon the buoyant force, which acts upward, and whose magnitude is
[tex]B=\rho_p V g[/tex]
where
[tex]\rho_p = 800 kg/m^3[/tex] is the density of paraffin
V is the volume of paraffin displaced, which corresponds to the volume of the block
[tex]g=10 m/s^2[/tex]
Substituting,
[tex]B=(800)(7.8\cdot 10^{-4})(10)=6.2 N[/tex]
Therefore, the apparent weight of the block in paraffin will be:
[tex]W'=W-B=70-6.1=63.9 N[/tex]
Answer:
63.788 N
Explanation:
Density of a metal block = 9000 ×〖10〗^(-6) kgm^(-3)
Mass of the metal block = 70/10 = 7 kg
Volume = Mass/ Density
Volume of the metal block = 7/9000 ×(10)^(-6) kgm^(-3)
= 777.777 m^3
When it’s immersed in water up thrust is acted on a metal block then the weight is reduced due to up thrust, Let’s take W as a New measurement,
W + U = 70
W = 70-U
W = 70 -Vρ_l g
W= 70 – 777.77×800×(10)^(-6)×10
W = 63.778 N
