Answer:
B false it is illegal to only have got fog lights on though and bright headlights because it can distract other drivers going last and if the y are distracted then that will cause a collision
Hope this helps :)
Explanation:
Answer:
Activation energy for creep in this temperature range is Q = 252.2 kJ/mol
Explanation:
To calculate the creep rate at a particular temperature
creep rate, ![\zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )](https://tex.z-dn.net/?f=%5Czeta_%7B%5Ctheta%7D%20%3D%20C%20%5Cexp%28%5Cfrac%7B-Q%7D%7BR%20%5Ctheta%7D%20%29)
Creep rate at 800⁰C, ![\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )](https://tex.z-dn.net/?f=%5Czeta_%7B800%7D%20%3D%20C%20%5Cexp%28%5Cfrac%7B-Q%7D%7BR%20%28800%2B273%29%7D%20%29)
![\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\](https://tex.z-dn.net/?f=%5Czeta_%7B800%7D%20%3D%20C%20%5Cexp%28%5Cfrac%7B-Q%7D%7B1073R%7D%20%29%5C%5C%5Czeta_%7B800%7D%20%3D%201%20%5C%25%20per%20hour%20%3D0.01%5C%5C)
.........................(1)
Creep rate at 700⁰C
![\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )](https://tex.z-dn.net/?f=%5Czeta_%7B700%7D%20%3D%20C%20%5Cexp%28%5Cfrac%7B-Q%7D%7BR%20%28700%2B273%29%7D%20%29)
![\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2} \% per hour =5.5 * 10^{-4}](https://tex.z-dn.net/?f=%5Czeta_%7B800%7D%20%3D%20C%20%5Cexp%28%5Cfrac%7B-Q%7D%7B973R%7D%20%29%5C%5C%5Czeta_%7B800%7D%20%3D%205.5%20%2A%2010%5E%7B-2%7D%20%20%5C%25%20per%20hour%20%3D5.5%20%2A%2010%5E%7B-4%7D)
.................(2)
Divide equation (1) by equation (2)
![\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\](https://tex.z-dn.net/?f=%5Cfrac%7B0.01%7D%7B5.5%20%2A%2010%5E%7B-4%7D%20%7D%20%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073R%7D%20-%5Cfrac%7B-Q%7D%7B973R%7D%20%5D%5C%5C18.182%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073R%7D%20%2B%5Cfrac%7BQ%7D%7B973R%7D%20%5D%5C%5CR%20%3D%208.314%5C%5C18.182%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073%2A8.314%7D%20%2B%5Cfrac%7BQ%7D%7B973%2A8.314%7D%20%5D%5C%5C18.182%3D%20%5Cexp%5B0.0000115%20Q%5D%5C%5C)
Take the natural log of both sides
![ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol](https://tex.z-dn.net/?f=ln%2018.182%3D%200.0000115Q%5C%5C2.9004%20%3D%200.0000115Q%5C%5CQ%20%3D%202.9004%2F0.0000115%5C%5CQ%20%3D%20252211.49%20J%2Fmol%5C%5CQ%20%3D%20252.2%20kJ%2Fmol)
Answer:
The heat of the arc melts the surface of the base metal and the end of the electrode. The electric arc has a temperature that ranges from 3,000 to 20,000 °C
Explanation:
Welding fumes are complex mixtures of particles and ionized gases.
Answer:
Explanation:
Pie charts generally should have no more than eight segments.
Answer:
0.08kg/s
Explanation:
For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.
The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.
finally you use the two previous equations to make a system and find the mass flows
I attached procedure