Respuesta :
Answer:
The kinetic increases by 48.84 %
Explanation:
The expression for the kinetic energy is:-
[tex]K.E.=\frac{1}{2}\times mv^2[/tex]
Where, m is the mass of the object
v is the velocity of the object
Let the new velocity is:- v'
v is increased by 22 %. Thus, v' = 1.22 v
So, the new kinetic energy is:-
[tex]K.E.'=\frac{1}{2}\times mv'^2=\frac{1}{2}\times m{1.22}^2=\frac{1}{2}\times mv^21.4884=1.4884K.E.[/tex]
Thus, the kinetic increases by 48.84 %
Answer:
the kinetic energy of the vehicle increases 48.84% after increasing its initial speed by 22%.
Explanation:
- KE = (1/2)mv²
∴ KE1 = (1/2)mv²
⇒ KE1 = (0.5)mv²
∴ KE2 = (1/2)m(v + 0.22v)²
⇒ KE2 = (1/2)m(1.22v)²
⇒ KE2 = (1/2)m(1.4884v²)
⇒ KE2/KE1 = ((1/2)m(1.4884v²))/((1/2)mv²)
⇒ KE2/KE1 = 1.4884 = 1 + 0.4884
⇒ % increase = (0.4884)×100 = 48.84%
