Answer to Question #145203 in Physics for Harrison

Question #145203
A solid cylinder has a density of 8.9g/cm^3 and weighs 72g. Find its apparent weight when immersed totally in a liquid of density 0.8g/cm^3.
1
Expert's answer
2020-11-18T14:52:13-0500

By the definition of the apparent weight, we have:


WA=WcFB,W_{A}=W_{c}-F_B,

here, WAW_{A} is the apparent weight of the solid cylinder, WcW_c is the weight of the cylinder in air, FB=ρliquidVliquidgF_B=\rho_{liquid}V_{liquid}g is the buoyant force that acts on the solid cylinder when it submerged in liquid, ρliquid=800 kgm3\rho_{liquid}=800\ \dfrac{kg}{m^3} is the density of the liquid, Vliquid=VcV_{liquid}=V_{c} is the volume of the liquid displaced that is equal to the volume of the solid cylinder and g=9.8 ms2g=9.8\ \dfrac{m}{s^2} is the acceleration due to gravity.

Let's first find the volume of the liquid displaced:


Vliquid=Vc=mρ,V_{liquid}=V_{c}=\dfrac{m}{\rho},Vliquid=0.072 kg8900 kgm3=8.1106 m3.V_{liquid}=\dfrac{0.072\ kg}{8900\ \dfrac{kg}{m^3}}=8.1\cdot 10^{-6}\ m^3.

Finally, we can find the apparent weight of the solid cylinder:


WA=WcFB,W_{A}=W_{c}-F_B,WA=mgρliquidVliquidg,W_{A}=mg-\rho_{liquid}V_{liquid}g,

WA=0.072 kg9.8 ms2800 kgm38.1106 m39.8 ms2=0.64 N.W_{A}=0.072\ kg\cdot 9.8\ \dfrac{m}{s^2}-800\ \dfrac{kg}{m^3}\cdot 8.1\cdot 10^{-6}\ m^3\cdot 9.8\ \dfrac{m}{s^2}=0.64\ N.

Answer:

WA=0.64 N.W_{A}=0.64\ N.


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