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V=π*r²*h=π*5²*3=75*π cm³
objętość kropli
Vk=(4/3)*π*r³
r=1mm=0,1cm
Vk=(4/3)*π*0,1³=1,33*π*10^-3 cm³
n=75*π/(1,33*π*10^-3)=56391
Pole podstawy = π*5² = 25π
V naczynia =25π*3 = 75π cm³
2.
1mm = 0,1 cm
V kuli = 4/3πr³
4/3π*(0,1)³= 4/3π*0,001 cm³
3.
75π÷(4/3π*0,001) = 75 ÷ (4/3*1/1000) = 75 ÷ 4/3000 = 75*3000/4 = 56250.
r = 1/2*10 = 5cm
V = πr²*h = π*5²*3 = 75π
V = 75πcm³ => 75000πmm³
Kula:
r = 1/2*2mm = 1mm
V = 4/3πr³ = 4/3πmm³
V = 4/3πmm³
Rozwiązanie:
75000π : 4/3π = 75000*3/4 = 225000/4 = 56250
Odp: Pojemnik napełni 56250 kropel wody.