Question

nos momentos de uma corrida um corredor de 60 kg de massa adquiriu uma energia cinética de 6750 J, calcule a velocidade do atleta nesse instante

Answer 1

Ec= $\frac{m v^{2} }{2}$

6750= $\frac{60 v^{2} }{2}$

6750 x 2 = 60v²
13500=60v²
v²=225

v= 15 m/s

Answer 2

Olá, boa noite ☺

Resolução.

$\\ \boxed{\texttt{Ec}= \dfrac{\texttt{m}\cdot \texttt{v}^{\text{2}}}{2} }\rightarrow \texttt{F\'{o}rmula.} \\ \\ \\ \texttt{Nomenclatura:} \\ \\ \texttt{Ec}= \texttt{energia cin\'{e}tica} \\ \\ \texttt{m=massa} \\ \\ \texttt{v=velocidade} \\ \\ \texttt{Aplicando valores:} \\ \\ \texttt{6750}= \dfrac{\texttt{60}\cdot \texttt{v}^{\text{2}}}{2} \\ \\ \texttt{60v}^{\tettt{2}} = \texttt{6750} \cdot \exttt{2} \\ \\ \texttt{60v}^{\tettt{2}} = \texttt{13500} \\ \\ \texttt{v}^{\texttt{2}}= \dfrac{\texttt{13500}}{\texttt{60}} \\ \\ \texttt{v}^{\texttt{2}}=225 \\ \\ \texttt{v}= \sqrt{\texttt{225}} \\ \\ \texttt{v}= \texttt{15 m/s}$

Bons estudos :)