Respuesta :
Given Information:
Mass = m = 5 kg
Initial velocity = v₁ = 200 m/s
Final velocity = v₂ = 150 m/s
Required Information:
change in thermal energy = ?
Answer:
ΔKE = 43,750 J
Explanation:
According to the principle of conservation of mechanical energy
PE₁ + KE₁ = PE₂ + KE₂
The projectile is fired from ground level so there is zero energy stored due tot he position of the projectile which means that the change in potential energy is zero, so the equation becomes
ΔKE = KE₁ - KE₂
ΔKE = ½mv₁² - ½mv₂²
ΔKE = ½m(v₁² - v₂²)
ΔKE = ½*5(200² - 150²)
ΔKE = 43,750 J
Therefore, the change in the thermal energy of the projectile and air is 43,750 Joules.
The mechanical energy is constant if the subject is an isolated system pertaining to conservative forces. The change in the thermal energy is 43,750 J.
Given that:
Mass = 5 kg
Initial velocity = 200 m/s
Final velocity = 150 m/s
Change in thermal energy = ?
The principle of conservation of mechanical energy states that it is equivalent to the sum of kinetic energies and potential energies in an isolated system, such that:
PE₁ + KE₁ = PE₂ + KE₂
As we know, the projectile fired from the ground level. The potential energy is zero.
The equation becomes:
[tex]\Delta KE &= KE_1 - KE_2\\\\\Delta KE &= \dfrac{1}{2} (mv_1)^2-\dfrac{1}{2} (mv_2)^2\\\\\Delta KE &= \dfrac{1}{2}[v_1^{(2)} - v_2^{(2)}]\\\\\Delta KE &= \dfrac{1}{2} \times 5 (200^2 - 150^2)[/tex]
[tex]\Delta KE &= 43,750 J[/tex]
Thus, the change in the thermal energy is 43,750 J.
To know more about thermal energy, refer to the following link:
https://brainly.com/question/11278589
