Question

Ajutati-ma cu exercitiul 4. Va rog.

Answer 1

[tex]\displaystyle \text{Se da:}\\ \\ R_p=6400km=6400000m\\ \\ M_p=6\times 10^{27}g=6\times 10^{24}kg\\ \\ K=6,67\times 10^{-11}\frac{N\times m^2}{kg^2}\\ \\ a)r=R_p,\text{ }v=?\frac ms\\ \\ b)r=R_p+h,\text{ }v=?\frac ms\\ \\ \\[/tex]


[tex]\displaystyle \text{Formule:}\\ \\ \text{Forta centripeta egaleaza forta atractiei universale:}\\ \\ \frac{m\times v^2}{r}=K\times\frac{m\times M_p}{r^2}\\ \\ v^2=K\times \frac{M_p}r\\ \\ v=\sqrt{K\times \frac {M_p}r}\\ \\ \\[/tex]


[tex]\displaystyle a)r=R_p\\ \\ v=\sqrt{K\times \frac {M_p}{R_p}}\\ \\ \\ b)r=R_p+h\\ \\ v=\sqrt{K\times \frac {M_p}{R_p+h}}\\ \\ \\ \text{Calcule:}\\ \\ a)v=\sqrt{6,67\times 10^{-11}\times \frac {6\times10^{24}}{6400000}}\approx7907,67\frac ms [/tex]


$\displaystyle \text{Avem o miscare accelerata schimbata}\\ \text{de aceea viteza o putem afla folosind acceleratie medie}\\ \\ \\ v=v_0+a_{med}\times t\\ \\ \\ \text{Acceleratie, la rindul ei, o putem afla din forta medie}\\ \\ \\ F_{med}=m\times a_{med},\text{ legea a 2-a a lui Newton}\\ \\ a_{med}=\frac{F_{med}}{m}\\ \\ \\ \text{Forta o putem afla din media aritmetica a fortei initiale si finale}$


[tex]\displaystyle F_{med}=\frac{F_0+F}2\\ \\ \text{Iar deoarece avem }F_0=0\text{ si }F=2\times t\text{, obtinem:}\\ \\ F_{med}=\frac{2\times t}{2}\\ \\ F_{med}=t\Rightarrow a_{med}=\frac tm\Rightarrow v=v_0+\frac {t^2}m\\ \\ \\ \text{Inlocuim in formula lucrului si exprimam formula finala:}\\ \\ \\[/tex]


[tex]\displaystyle L=\frac m2\times[(v_0+\frac {t^2}m)^2-v_0^2]\\ \\ L=\frac m2\times(v_0^2+\frac{2\times v_0\times t^2}m+\frac{t^4}{m^2}-v_0^2)\\ \\ L=\frac m2\times(\frac{2\times v_0\times t^2}m+\frac{t^4}{m^2})\\ \\ L=\frac {2\times v_0\times t^2\times m}{2\times m}+\frac{m\times t^4}{2\times m^2}\\ \\ L=v_0\times t^2+\frac{t^4}{2\times m}\\ \\ \\ \boxed{L=t^2\times(v_0+\frac {t^2}{2\times m})}\\ \\ \\ \text{Calcule:}\\ \\ L=10^2\times(6+\frac{10^2}{2\times 1})=5600J[/tex]