## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 9 - Problems - Page 366: 89

#### Answer

(a) The weight of the block is $0.794~N$ (b) The scale reading is $0.544~N$

#### Work Step by Step

(a) We can find the mass $M$ of the block: $M = \rho~V$ $M = (2700~kg/m^3)(0.020~m)(0.030~m)(0.050~m)$ $M = 0.081~kg$ We can find the weight of the block: $weight = Mg = (0.081~kg)(9.80~m/s^2) = 0.794~N$ The weight of the block is $0.794~N$ (b) According to Archimedes' principle, the buoyant force is equal to the weight of the displaced oil. We can find the buoyant force: $F_b = m_o~g$ $F_b = \rho_o~V~g$ $F_b = (850~kg/m^3)(3.0\times 10^{-5}~m^3)(9.80~m/s^2)$ $F_b = 0.250~N$ We can find the scale reading when the block is submerged in oil: $F_{scale} = 0.794~N-0.250~N = 0.544~N$ The scale reading is $0.544~N$

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