Jawaban:
[tex] \huge { \blue {2 \ Joule}} [/tex]
Penjelasan:
massa (m) = 1 kg
kecepatan awal ([tex] v_{0} [/tex]) = 4 m/s
kecepatan akhir ([tex] v_{1} [/tex]) = 8 m/s
Perubahan energi kinetik (Ek)?
Rumus energi kinetik: [tex] \boxed {Ek = \frac{1}{2} \times m \times {v}^{2} } [/tex], maka:
Perubahan energi kinetik
=> [tex] Ek_{1} - Ek_{0} \\ ( \frac{1}{2} \times m \times v_{1}) - ( \frac{1}{2} \times m \times v_{0}) \\ ( \frac{1}{2} \times 1 \times 8) - ( \frac{1}{2} \times 1 \times 4) \\ ( \frac{8}{2} ) - ( \frac{4}{2}) \\ 4 - 2 \\ \boxed{2 \: Joule} [/tex]
Jadi, perubahan energi kinetik bila bola dipercepat adalah [tex] \blue {2 \ Joule} [/tex].
[tex] \large {\boxed {\blue {\star \:Answered \: By: \: \bold {sulkifli2018} \star} } } [/tex]
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Verified answer
Jawaban:
[tex] \huge { \blue {2 \ Joule}} [/tex]
Penjelasan:
Diketahui:
massa (m) = 1 kg
kecepatan awal ([tex] v_{0} [/tex]) = 4 m/s
kecepatan akhir ([tex] v_{1} [/tex]) = 8 m/s
Ditanyakan:
Perubahan energi kinetik (Ek)?
Penyelesaian:
Rumus energi kinetik: [tex] \boxed {Ek = \frac{1}{2} \times m \times {v}^{2} } [/tex], maka:
Perubahan energi kinetik
=> [tex] Ek_{1} - Ek_{0} \\ ( \frac{1}{2} \times m \times v_{1}) - ( \frac{1}{2} \times m \times v_{0}) \\ ( \frac{1}{2} \times 1 \times 8) - ( \frac{1}{2} \times 1 \times 4) \\ ( \frac{8}{2} ) - ( \frac{4}{2}) \\ 4 - 2 \\ \boxed{2 \: Joule} [/tex]
Jadi, perubahan energi kinetik bila bola dipercepat adalah [tex] \blue {2 \ Joule} [/tex].
[tex] \large {\boxed {\blue {\star \:Answered \: By: \: \bold {sulkifli2018} \star} } } [/tex]