Question

A 30 gram bullet travels at 300 m/s. How much kinetic energy does it have?​

Answer 1

Answer:

1.35 kJ  

Explanation:

KE = ½mv²

     = ½ × 0.030  kg × (300 m·s⁻¹)²

      = 1350 J

      = 1.35 kJ

Answer 2

It will contain kinetic energy of 1.35kJ.

Mass (m) of the bullet$= 30 \ grams$

The velocity of the bullet (v) $= 300\ m/s$

The kinetic energy of the bullet, we can find out using the formula,

$E_k = \frac{1}{2} mv^2$

where,

Eₖ = Kinetic energy,

m = Mass

v = Velocity.

Hence,

$Eₖ = \frac{1}{2} (30 g)(300 m/s)^2\\\\= (15 g)(90000 m^2/s^2)\\= 1350000 g m^2/s^2\\\\\\\\= 1350 kg m^2/s^2$

$= 1350 \ J$

$= 1.35\ kJ$

Hence, the Kinetic Energy is 1.35 kJ.

#SPJ2