Respuesta :
- Two 3.0g bullets are fired with speeds of 40.0 m/s and 80.0 m/s, respectively.
- Mass of the 1st bullet (m1) = 3 g = 0.003 Kg
- Mass of the 2nd bullet (m2) = 3 g = 0.003 Kg
- Velocity of the 1st bullet (v1) = 40 m/s
- Velocity of the 2nd bullet (v2) = 80 m/s
- We know, kinetic energy of a body
[tex] = \sf \frac{1}{2} m {v}^{2} [/tex]
- Kinetic energy of the 1st bullet [tex] \sf = \frac{1}{2} \times m1 \times {(v1)}^{2} \\ \sf= \frac{1}{2} \times 0.003 \times {(40)}^{2}J \\ \sf = \frac{1}{2} \times 0.003 \times 1600J \\ \sf = 2.4J[/tex]
- Kinetic energy of the 2nd bullet [tex] \sf = \frac{1}{2} \times m2 \times {(v2)}^{2} \\ \sf= \frac{1}{2} \times 0.003 \times {(80)}^{2}J \\ \sf = \frac{1}{2} \times 0.003 \times 6400J \\ \sf = 9.6J[/tex]
- So, the 2nd bullet which has greater velocity has more kinetic energy.
- Therefore, the ratio of their kinetic energies
[tex] \sf = \frac{2.4J}{9.6J} \\ \sf= \frac{24}{96} = \frac{1}{4} \\ \sf = 1 : 4[/tex]
Answer:
The kinetic energies of the bullets are 2.4 J and 9.6 J.
The bullet having greater velocity has more kinetic energy.
The ratio of their kinetic energies is 1 : 4.
Hope you could get an idea from here
Doubt clarification - use comment section.
The kinetic energy of the two bullets are 2.4 J and 9.6 J respectively.
The ratio of the kinetic energy of the bullets is 1:4.
What is kinetic energy?
The kinetic energy of an object is the energy possessed by the object due to its motion.
The kinetic energy of the two bullets is calculated as follows;
[tex]K.E_1 = \frac{1}{2} \times 0.003 \times 40^2 = 2.4 \ J\\\\K.E_2 = \frac{1}{2} \times 0.003 \times 80^2 = 9.6 \ J[/tex]
Ratio of the kinetic energy of the bullets is calculated as follows;
[tex]K.E_1 : K.E_2 = 2.4: 9.6 \ = \ 1: 4[/tex]
Learn more about kinetic energy here: https://brainly.com/question/25959744
