Physics, asked by zakakareem1255, 1 year ago

A piece of iron of density is 7.8 into 10 raise to the power 3 kg per metre cube and volume 100 cm cube is totally immersed in water. Calculate the weight of iron piece in air, upthrust and apparent weight in water

Answers

Answered by Brainlycutipie
5


Density of iron ρ = 7.8 x 103 kg/m3

Volume of iron piece V = 100 cm3 = 100 x 10−6 m3

Mass of the iron piece M = Vρ = 100 x 10−6 m3 x 7.8 x 103 kg/m3
                                             = 0.78 kg
(i) Weight of the iron piece in air = Mg = 0.78 kg x 9.8 m/s2
                                                    = 7.644 kg. m /s2 = 7.644 N

(ii) Upthrust = Volume of the solid x density of water x g
                  = Vρwg
                  = 100 x 10−6 m3 x 1000 kg/m3 x 9.8 m/s2
                  = 0.98 N
(iii) Apparent weight of iron piece in water = Weight in air − upthrust
                                                            =  7.644 N − 0.98 N
                                                            =  6.664 N





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