exerted on the area directly under the cube before melting to the pressure P2cap P sub 2
P1=miga2=ρia3ga2=ρiagcap P sub 1 equals the fraction with numerator m sub i g and denominator a squared end-fraction equals the fraction with numerator rho sub i a cubed g and denominator a squared end-fraction equals rho sub i a g ρirho sub i is the density of ice. After melting, the mass of water equals the mass of ice . The water spreads over the entire bottom area olimpiadnye zadachi s reshenijami po fizike 8 klass
Heat lost by hot water must equal heat gained by the melting ice and the resulting cold water: exerted on the area directly under the cube
Physics olympiads for 8th-grade students (Grade 8) typically cover mechanics (kinematics, statics, pressure), thermal phenomena (calorimetry, phase transitions), and basic DC circuits. Below are representative problems and solutions based on tasks from the All-Russian Physics Olympiad (VOS) and the Sirius Educational Center . Task: An ice cube with edge length Below are representative problems and solutions based on
Before melting, the pressure directly under the cube is caused by its weight migm sub i g
η=AusefulAtotal⋅100%=67.5⋅100%=80%eta equals the fraction with numerator cap A sub useful end-sub and denominator cap A sub total end-sub end-fraction center dot 100 % equals 6 over 7.5 end-fraction center dot 100 % equals 80 % The efficiency of the block system is 80%.