[latex]V=frac{4}{3}pi r^3 = [/latex] [latex]V=frac{4}{3}pi 3^3=[/latex] [latex]V=frac{4}{3}pi 1331 =[/latex] [latex]V= 1331 * frac{4}{3}pi=1774.67 pi cm^3[/latex] [latex]V=frac{4}{3}*pi*1331=5572.45 cm^3[/latex] 1kg=100dag 100dag=1000g [latex]p=1frac{g}{cm^3}[/latex] [latex]5572.45*1frac{g}{cm}^3+430g=5572g+430g= 6002.45g[/latex] [latex]6002.45g=6.00245kg[/latex] [latex]p=7.9frac{g}{cm^}[/latex] [latex]5572.45*7.9frac{g}{cm}^3+430g=44022.36g+430g=44452.36g[/latex] [latex] 44452.36g=44.45236kg[/latex] Myślę, że pomogłem :)
dane: M = 430 g - masa niewypełnionej piłki d = 22 cm r = d/2 = 11 cm p1 = 1 g/cm³ - gęstość wody p2 = 7,9 g/cm³ - gęstość żelaza szukane: m1, m2 = ? d = m/V I*V m = M + V*d V = 4/3 πr³ = 4/3 *3,14*(11cm)³ = 5 572,45 cm³ m = M + V*d m1 = M + V*d1 = 430g + 5572,45cm³*1g/cm³ = 430g + 5572,45g = 6 002,45 g ≈ 6 kg m1 ≈ 6 kg m2 = M + V*d2 = 430g + 5572,45cm³*7,9g/cm³ = 430g + 44022,36g = 44 452,36 g ≈ 44,5 kg m2 ≈ 44,5 kg
