Answer:
A. 28.42 m/s
B. 41.21 m.
Explanation:
A. Determination of the initial velocity of the ball:
Time (t) to reach the maximum height = 2.9 s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) =?
Thus, we can obtain the initial velocity of the ball as follow:
v = u + gt
0 = u + (–9.8 × 2.9)
0 = u – 28.42
Collect like terms
u = 0 + 28.42
u = 28.42 m/s
Therefore, the initial velocity of the ball is 28.42 m/s.
B. Determination of the maximum height reached.
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) = 28.42 m/s.
Maximum height (h) =?
Thus, we can obtain the maximum height reached by the ball as follow:
v² = u² + 2gh
0² = 28.42² + (2 × –9.8 × h)
0 = 807.6964 + (–19.6h)
0 = 807.6964 – 19.6h
Collect like terms
0 – 807.6964 = – 19.6h
– 807.6964 = – 19.6h
Divide both side by – 19.6
h = –807.6964 / –19.6
h = 41.21 m
Therefore, the maximum height reached by the ball is 41.21 m
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As they evolved and adapted, those of the fittest survived to reproduce. Those that did not adapt, died.
Answer:
snow
Explanation:
Since the process undergoes adiabatic expansion, hence q = 0 and ΔU = w.
We can sole this problem using the following derivation:
![ln(\frac{T_2}{T_1} )=-(\gamma -1)ln(\frac{V_f}{V_i} )=-(\gamma -1)ln(\frac{T_2}{T_1}\frac{P_i}{P_f} )\\=-(\gamma -1)ln(\frac{T_2}{T_1})-(\gamma -1)ln(\frac{P_i}{P_f})\\=-(\frac{\gamma -1}{\gamma})ln(\frac{P_i}{P_f})\\=-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\ln(\frac{T_2}{T_1} )==-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\Substituting\ values:\\\\](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BT_2%7D%7BT_1%7D%20%29%3D-%28%5Cgamma%20-1%29ln%28%5Cfrac%7BV_f%7D%7BV_i%7D%20%29%3D-%28%5Cgamma%20-1%29ln%28%5Cfrac%7BT_2%7D%7BT_1%7D%5Cfrac%7BP_i%7D%7BP_f%7D%20%20%29%5C%5C%3D-%28%5Cgamma%20-1%29ln%28%5Cfrac%7BT_2%7D%7BT_1%7D%29-%28%5Cgamma%20-1%29ln%28%5Cfrac%7BP_i%7D%7BP_f%7D%29%5C%5C%3D-%28%5Cfrac%7B%5Cgamma%20-1%7D%7B%5Cgamma%7D%29ln%28%5Cfrac%7BP_i%7D%7BP_f%7D%29%5C%5C%3D-%28%5Cfrac%7B%5Cfrac%7BC_%7Bp%2Cm%7D%7D%7BC_%7Bp%2Cm%7D-R%7D%20-1%7D%7B%5Cfrac%7BC_%7Bp%2Cm%7D%7D%7BC_%7Bp%2Cm%7D-R%7D%7D%29ln%28%5Cfrac%7BP_i%7D%7BP_f%7D%29%5C%5C%5C%5Cln%28%5Cfrac%7BT_2%7D%7BT_1%7D%20%29%3D%3D-%28%5Cfrac%7B%5Cfrac%7BC_%7Bp%2Cm%7D%7D%7BC_%7Bp%2Cm%7D-R%7D%20-1%7D%7B%5Cfrac%7BC_%7Bp%2Cm%7D%7D%7BC_%7Bp%2Cm%7D-R%7D%7D%29ln%28%5Cfrac%7BP_i%7D%7BP_f%7D%29%5C%5C%5C%5CSubstituting%5C%20values%3A%5C%5C%5C%5C)
![ln(\frac{T_2}{T_1} )=-(\frac{\frac{28.86}{28.86-8.314} -1}{\frac{28.86}{28.86-8.314}})ln(\frac{0.802\ atm}{0.602\ atm})=-0.0826\\\\ln(\frac{T_2}{T_1} )=-0.0826\\\\Taking\ exponential\ of\ both \ sides:\\\\\frac{T_2}{T_1} =e^{-0.0826}\\\\T_2=0.9207T_1\\\\T_2=0.9207*288\\\\T_2=265\ K\\](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BT_2%7D%7BT_1%7D%20%29%3D-%28%5Cfrac%7B%5Cfrac%7B28.86%7D%7B28.86-8.314%7D%20-1%7D%7B%5Cfrac%7B28.86%7D%7B28.86-8.314%7D%7D%29ln%28%5Cfrac%7B0.802%5C%20atm%7D%7B0.602%5C%20atm%7D%29%3D-0.0826%5C%5C%5C%5Cln%28%5Cfrac%7BT_2%7D%7BT_1%7D%20%29%3D-0.0826%5C%5C%5C%5CTaking%5C%20exponential%5C%20of%5C%20both%20%5C%20sides%3A%5C%5C%5C%5C%5Cfrac%7BT_2%7D%7BT_1%7D%20%3De%5E%7B-0.0826%7D%5C%5C%5C%5CT_2%3D0.9207T_1%5C%5C%5C%5CT_2%3D0.9207%2A288%5C%5C%5C%5CT_2%3D265%5C%20K%5C%5C)
Since T2 = 265 K, we should expect a snow