Question

Orice exercitiu rezolvat de acolo e indeajuns, ofer coronita si puncte ​

Answer 1

Ti-am facut primele 3 probleme, mai mult de atat nu pot. Nu sunt 100% sigur ca le-am rezolvat corect pentru ca au trecut 2 ani de cand am lucrat la mecanica, dar sper sa te ajute.

Problema 7

d = 60 m

t = 10 s

---------------------------------

a) a = ?

b) v₁ = ?

c) Δd = ? (a₁ = -9 m/s²)

----------------------------------

a) $\displaystyle{ v_{0} = 0 \ m/s }$

$\displaystyle{ v = \frac{d}{t} = \frac{60}{10} = 6 \ m/s }$

$\displaystyle{ legea \ vitezei \ pentru \ MRUV \rightarrow v = v_{0} + a \cdot t }$

$\displaystyle{ \rightarrow a = \frac{v - v_{0} }{t} }$

$\displaystyle{ a = \frac{6}{10} = 0,6 \ m/s^{2} }$

b) $\displaystyle{ ecuatia \ lui \ Galilei \rightarrow v_{1}^{2} = v_{0}^{2} + 2 \cdot a \cdot d }$

$\displaystyle{ v_{1}^{2} = 2ad = 2 \cdot 0,6 \cdot 60 }$

$\displaystyle{ v_{1}^{2} = 1,2 \cdot 60 = 72 \rightarrow v_{1} = \sqrt{72} = 8,48 \ m/s }$

c) $\displaystyle{ legea \ miscarii \ pentru \ MRUV \rightarrow \Delta d = | d + \frac{a_{1}t^{2}}{2}| }$

$\displaystyle{ \Delta d = |60 + \frac{-9 \cdot 10^{2}}{2} | = |60 + \frac{-900}{2} | }$

$\displaystyle{ \Delta d = |60 - 450| = |-390| = 390 \ m }$

Deci sania a mai parcurs 390 metri dupa terminarea dealului, apoi s-a oprit.

Problema 8

$\displaystyle{ v_{0} = 4 \ m/s }$

x = 400 m

t = 40 s

----------------------------------------------------

a) a = ?

b) v = ?

c) $\displaystyle{ v_{med} = ? \ (x_{1} = 93,75 \ m) }$

------------------------------------------------------

a) $\displaystyle{ v = \frac{x}{t} = \frac{400}{40} = 10 \ m/s }$

$\displaystyle{ legea \ vitezei \ pentru \ MRUV \rightarrow v = v_{0} + a\cdot t }$

$\displaystyle{ \righarrow a = \frac{v-v_{0}}{t} = \frac{10-4}{40}= \frac{6}{40}^{(2} }$

$\displaystyle{ a = \frac{3}{20} = 0,15 \ m/s^{2} }$

b) am calculat mai sus

c) $\displaystyle{ la \ x_{1} = 93,75 \ m, \ v_{2} = \frac{93,75}{40} = 2,34 \ m/s }$

$\displaystyle{ v_{med} = \frac{v_{1}+v_{2}}{2} = \frac{2,34+10}{2} }$

$\displaystyle{ v_{med} = \frac{12,34}{2} = 6,17 \ m/s }$

Problema 9

$\displaystyle{ v_{0} = 500 \ m/s }$

v = 300 m/s

d = 0,2 m

--------------------------------------

a) a = ?

b) t = ?

c) $\displaystyle{ d_{max} = ? }$

-----------------------------------------

a) $\displaystyle{ ecuatia \ lui \ Galilei \rightarrow v_{0}^{2} = v^{2} + 2 \cdot a \cdot d }$

$\displaystyle{ a = \frac{v_{0}^{2} - v^{2}}{2d} = \frac{500^{2}-300^{2}}{2\cdot 0,2} }$

$\displaystyle{ a = \frac{250.000-90.000}{0,4} = \frac{160.000}{0,4} }$

$\displaystyle{ a = 400.000 \ m/s^{2} = 4 \cdot 10^{5} \ m/s^{2} }$

b) $\displaystyle{ v_{0} = v + a\cdot t \rightarrow t = \frac{v_{0}-v}{a} }$

$\displaystyle{ t = \frac{500-300}{400.000} = \frac{200}{400.000} }$

$\displaystyle{ t = 0,0005 \ s = 5 \cdot 10^{-4} \ s }$

c) Daca blindajul are valoare maxima inseamna ca proiectilul nu poate strapunge mai mult decat acea grosime. Deci la sfarsitul blindajului, viteza proiectilului va fi zero.

$\displaystyle{ d = max \rightarrow v = 0 \ m/s }$

$\displaystyle{ v_{0}^{2} = 2 \cdot a \cdot d_{max} \rightarrow d_{max} = \frac{v_{0}^{2}}{2a} }$

$\displaystyle{ d_{max} = \frac{500^{2}}{2 \cdot 4 \cdot 10^{5}} = \frac{2,5 \cdot 10^{5}}{8 \cdot 10^{5}} }$

$\displaystyle{ d_{max} = \frac{2,5}{8} = 0,3125 \ m }$