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Question Number 87179 by peter frank last updated on 03/Apr/20

Commented by peter frank last updated on 03/Apr/20

$$qn2$$

Commented by Rio Michael last updated on 03/Apr/20

$$2.\mathrm{volume}\mathrm{of}\mathrm{iron}=\frac{m}{\rho }=\frac{360\mathrm{g}}{6{\mathrm{gcm}}^{-3}}=60{\mathrm{cm}}^3$$$$\Rightarrow \mathrm{volume}\mathrm{of}\mathrm{submerge}=30{\mathrm{cm}}^3$$$$\mathrm{but}\mathrm{T}+{\mathrm{F}}_{upthrust}=\mathrm{W}$$$$\Rightarrow T=\mathrm{mg}-{\mathrm{F}}_{upthrust}=0.36\mathrm{kg}\times 9.{8\mathrm{ms}}^{-2}-30\times {10}^{-6}{\mathrm{m}}^3\times 6000{\mathrm{kgm}}^{-3}\times 9.8$$$$\mathrm{T}=3.528-1.764=1.764\mathrm{N}$$

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