What is the potential energy of a 2500 g object suspended 5 m above the earth's surface?
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Answer:
0.0613°C
Explanation:
the given parameters are m=15gm=15×10⁻³ V₁=865m/s V₂=534m/s
the bullet moves with different kinetic energies before and after the penetration, therefore
Kinetic energy before - kinetic energy after = 1/2 × m × ( V₁² - V₂²)
= × 15×10⁻³ × (865² - 534²)
= 3.47 × 10⁻³J
this loss in energy is transferred to the water, therefore
change in temperature =
where c = heat capacity of water = 4.19 x 10^3
m = mass of water = 13.5 kg
= {3.47 × 10⁻³} / {13.5 x 4.19 x 10^3 }
=0.0613°C