Answer:
The total hole mobility is 41.67 cm²/V s
Explanation:
Data given by the exercise:
hole mobility due to lattice scattering = ul = 50 cm²/V s
hole mobility due to ionized impurity = ui = 250 cm²/V s
The total mobility is equal:
Answer:
Explanation:
Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.
Low angle grain boundaries (quasi-coherent) are formed by the dislocation network positioned along the geometric plane with small tilt angle differences between successive peers that is tilt boundary made up edge dislocations therefore it may only divert the slip direction of the incoming gliding dislocation with very little frictional stresses. And on the other hand, a high angle grain boundary region because of their disordered almost liquid like structure which acts as a strong barrier against dislocation slip motion and causes actually formation of dislocations file-up against it by arresting their motion unless that the stress concentration at the leading dislocation becomes high enough to go though the barrier.
Answer:
20, 083 L
Explanation:
The mistake was the result of not using units when converting the 7862 l to Kg. They used the density in pounds hence they multiplied by 1.77 Lb/L and obtained 13597 Lb not Kg as they assumed.
To obtain the amount needed to refuel they subtracted this quantity from the 22,300 Kg required for the trip again obtaining the wrong quantity of 8703 Kg and they converted this to liters by dividing the density to get 4916 L and then placed then 5000 L of fuel
The quantity required was
7862 L * 1.77 Lb/L = 13915.74 Lb (pounds not kilos)
then converting this pounds to Kg by multiplying by 0.454 Kg/L one gets
6173 Kg on board
Amount Required
( 22,300 -6173) : 16127 Kg
16127 Kg/ 0.803 Kg/L = 20083 L
