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<u>Air pressure has no effect at all in an ideal gas approximation. This is because pressure and density both contribute to sound velocity equally, and in an ideal gas the two effects cancel out, leaving only the effect of temperature. Sound usually travels more slowly with greater altitude, due to reduced temperature.</u>
Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg
When you throw the ball in the air it is considered kinetic energy. Once the ball reaches its max height, it stops moving and all kinetic energy turns into potential energy. So when the ball is at its highest point.
Answer:
![\dfrac{K_t}{K_r}=\dfrac{5}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7BK_t%7D%7BK_r%7D%3D%5Cdfrac%7B5%7D%7B2%7D)
Explanation:
Given that,
Mass of the bowling ball, m = 5 kg
Radius of the ball, r = 11 cm = 0.11 m
Angular velocity with which the ball rolls, ![\omega=2.8\ rad/s](https://tex.z-dn.net/?f=%5Comega%3D2.8%5C%20rad%2Fs)
To find,
The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.
Solution,
The translational kinetic energy of the ball is :
![K_t=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=K_t%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
![K_t=\dfrac{1}{2}m(r\omega)^2](https://tex.z-dn.net/?f=K_t%3D%5Cdfrac%7B1%7D%7B2%7Dm%28r%5Comega%29%5E2)
![K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2](https://tex.z-dn.net/?f=K_t%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%205%5Ctimes%20%280.11%5Ctimes%202.8%29%5E2)
The rotational kinetic energy of the ball is :
![K_r=\dfrac{1}{2}I \omega^2](https://tex.z-dn.net/?f=K_r%3D%5Cdfrac%7B1%7D%7B2%7DI%20%5Comega%5E2)
![K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2](https://tex.z-dn.net/?f=K_r%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cdfrac%7B2%7D%7B5%7Dmr%5E2%5Ctimes%20%5Comega%5E2)
![K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2](https://tex.z-dn.net/?f=K_r%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cdfrac%7B2%7D%7B5%7D%5Ctimes%205%5Ctimes%20%280.11%29%5E2%5Ctimes%20%282.8%29%5E2)
Ratio of translational to the rotational kinetic energy as :
![\dfrac{K_t}{K_r}=\dfrac{5}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7BK_t%7D%7BK_r%7D%3D%5Cdfrac%7B5%7D%7B2%7D)
So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2