Length of time in minutes given time in seconds
minutes = seconds/60
Answer:
The Laplace transformation of the function would be
![F(s) = -\frac{1-2e^{-s}}{s}](https://tex.z-dn.net/?f=F%28s%29%20%3D%20-%5Cfrac%7B1-2e%5E%7B-s%7D%7D%7Bs%7D)
Step-by-step explanation:
According to the information of your problem
![F(s) = L\{f(t)\} = \int\limits_{0}^{\infty} e^{-st} f(t) dt](https://tex.z-dn.net/?f=F%28s%29%20%3D%20L%5C%7Bf%28t%29%5C%7D%20%3D%20%5Cint%5Climits_%7B0%7D%5E%7B%5Cinfty%7D%20%20e%5E%7B-st%7D%20f%28t%29%20dt)
And the function given is
![f(t) = \left \{ {{-1 \,\,\,\,\,\,\,0 \leq t < 1} \atop {1 \,\,\,\,\,\,\, t \geq 1}} \right.](https://tex.z-dn.net/?f=f%28t%29%20%3D%20%20%5Cleft%20%5C%7B%20%7B%7B-1%20%20%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C0%20%5Cleq%20t%20%3C%201%7D%20%5Catop%20%7B1%20%20%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%20t%20%5Cgeq%20%201%7D%7D%20%5Cright.)
Therefore
![F(s) = L\{f(t)\} = \int\limits_{0}^{1} -e^{-st} dt + \int\limits_{1}^{\infty} e^{-st} dt](https://tex.z-dn.net/?f=F%28s%29%20%3D%20L%5C%7Bf%28t%29%5C%7D%20%3D%20%5Cint%5Climits_%7B0%7D%5E%7B1%7D%20%20-e%5E%7B-st%7D%20%20dt%20%2B%20%5Cint%5Climits_%7B1%7D%5E%7B%5Cinfty%7D%20%20e%5E%7B-st%7D%20%20dt)
And when you compute those integrals you get that
![F(s) = -\frac{1-2e^{-s}}{s}](https://tex.z-dn.net/?f=F%28s%29%20%3D%20-%5Cfrac%7B1-2e%5E%7B-s%7D%7D%7Bs%7D)
Answer:
84.6
Step-by-step explanation:
135c + 273.15 = 408 K
72C + 273.15 = 345 K
100 L over 408 K = V2 over 345 K
V2= 84.6