this is a 3 part question20) A 9.50-g bullet has a speed of 1.30 km>s. (a) What is its kinetic energy in joules? (b) What is the bullet’s kinetic energy if its speed is halved? (c) If its speed is doubled?

Respuesta :

Given that mass of bullet, m = 9.50 g = 0.0095 kg

speed of bullet, v is 1.30 km/s

(a) Kinetic energy is given by the formula

[tex]K\mathrm{}E\text{. = }\frac{1}{2}mv^2[/tex]

Substituting the values in the above formula, we get

[tex]K\mathrm{}E\mathrm{}=\frac{1}{2}(0.0095)(1.30)^2=8.0275\times10^{-3}\text{ J}[/tex]

(b) Speed of bullet is v/2

Sustituting this value in the formula of kinetic energy, we get

[tex]\begin{gathered} K.E._1=\text{ }\frac{1}{2}m(\frac{v}{2})^2 \\ =\frac{1}{2}\times(0.0095)\times(\frac{1.30}{2})^2 \\ =2.0068\times10^{-3\text{ }}J \end{gathered}[/tex]

(c) Speed of bulllet becomes 2v

Sustituting this value in the formula of kinetic energy, we get

[tex]\begin{gathered} K.E._2=\text{ }\frac{1}{2}m(2v)^2 \\ =\frac{1}{2}\times(0.0095)\times(2\times1.30)^2 \\ =0.03211J \end{gathered}[/tex]