Question

Un cub cu latura=8cm produce o alungire cu 51,2 mm unui resort elastic cu k=270 N/m sa se afle densitatea

Answer 1

$a=8 \ cm=0,08 \ m\\\triangle l=51,2 \ mm=0,0512 \ m \\ k=270 \ N/m \\ g=10 \ N/kg \\ \rho-? \\ \rho= \frac{m}{V} \\ m= \frac{G}{g} \\ G=F_e \\ F_e=k \cdot \triangle l \\ F_e=270 \ N/m \cdot0,0512 \ m \\ F_e=13,824 \ N \\ \Rightarrow G=13,824 \ N \\ m= \frac{13,824 \ N}{10 \ N/kg} \\ m=1,3824 \ kg \\ V=a^3 \\ V=(0,08 \ m)^3 \\ V=0,000512 \ m^3 \\ \Rightarrow \rho= \frac{1,3824 \ kg}{0,000512 \ m^3 } \\ \Rightarrow \boxed{\rho=2700 \ kg/m^3=Aluminiu}.$

Answer 2

Δl=51,2 mm; k=270N/m;
⇒F=Δl·k
⇒F=51,2mm·270N/m
⇒F=0,0512m·270N/m⇒se reduce m[metru] cu m[metru];
⇒F=0,0512·270N
⇒F=13,824N;
⇒F=G⇒G=13,824N;
⇒G=m·g⇒g=10N/kg;
⇒m=G/g
⇒m=13,824N/10N/kg⇒se reduce N[newton] cu N[newton];
⇒m=13,824/10kg
⇒m=1,3824kg;
⇒l cub=8cm⇒V=l³⇒V=8³cm³⇒V=512cm³;
⇒p=m/V
⇒p=1,3824kg/512cm³
⇒p=0,0027kg/cm³=2700kg/m³;