Answer:
Explanation:
Ohms Law I=E/R (resistive requires no power factor correction)
150/25= 6 amps
Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as
![a(t)=5(1-\frac{t}{15})](https://tex.z-dn.net/?f=a%28t%29%3D5%281-%5Cfrac%7Bt%7D%7B15%7D%29)
![a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bd%5E%7B2%7Dx%7D%7Bdt%5E%7B2%7D%7D%5C%5C%5C%5C%5Ctherefore%20%5Cfrac%7Bd%5E%7B2%7Dx%7D%7Bdt%5E%7B2%7D%7D%3D5%281-%5Cfrac%7Bt%7D%7B15%7D%29)
Integrating both sides we get
![\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bd%5E%7B2%7Dx%7D%7Bdt%5E%7B2%7D%7Ddt%3D%5Cint%205%281-%5Cfrac%7Bt%7D%7B15%7D%29dt%5C%5C%5C%5C%5Cfrac%7Bdx%7D%7Bdt%7D%3Dv%3D5t-%5Cfrac%7B5t%5E%7B2%7D%7D%7B30%7D%2Bc)
Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get
![\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bdx%7D%7Bdt%7Ddt%3D%5Cint%20%285t-%5Cfrac%7B5t%5E%7B2%7D%7D%7B30%7D%29dt%5C%5C%5C%5Cx%28t%29%3D%5Cfrac%7B5t%5E%7B2%7D%7D%7B2%7D-%5Cfrac%7B5t%5E%7B3%7D%7D%7B90%7D%2BD)
Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of
![x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m](https://tex.z-dn.net/?f=x%2815%29%3D2.5%5Ctimes%2015%5E%7B2%7D-%5Cfarc%7B15%5E%7B3%7D%7D%7B18%7D%3D375m)
Thus the remaining 125 meters will be covered with a constant speed of
![v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s](https://tex.z-dn.net/?f=v%2815%29%3D5%5Ctimes%2015-%5Cfrac%7B15%5E%7B2%7D%7D%7B6%7D%3D37.5m%2Fs)
in time equalling ![t_{2}=\frac{125}{37.5}=3.33seconds](https://tex.z-dn.net/?f=t_%7B2%7D%3D%5Cfrac%7B125%7D%7B37.5%7D%3D3.33seconds)
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds
Answer:
(a) Flow rate of vehicles = No of vehicles per mile * Speed
=No of cars per mile * Speed +No of trucks per mile * Speed
= 0.75*50*60 + 0.25*50*40
=2750 vehicles / hour
(b) Let Density of vehicles on grade = x
Density on flat * Speed =Density on grade * Speed
So,( 0.75*50) * 60 + (0.25*50) * 40 = (0.75* x) * 55 + (0.25* x) * 25
So, x= 57.89
So, Density is around 58 Vehicles per Mile.
(c) Percentage of truck by aerial photo = 25%
(d)Percentage of truck bystationary observer on the grade= 25*30/60 * 25/55 =22.73 %
Answer:
(a) The Final Temperature is 315.25 K.
(b) The amount of mass that has entered 0.5742 Kg.
(c) The work done is 56.52 kJ.
(d) The entrophy generation is 0.0398 kJ/kgK.
Explanation:
Explanation is in the following attachments.