Answer:
0.546 ohm / μm
Explanation:
Given that :
N = 1.015 * 10^17
Electron mobility, u = 3900
Hole mobility, h = 1900
Ng = 4.42 x10^22
q = 1.6*10^-19
Resistivity = 1/qNu
Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)
= 0.01578880889 ohm /cm
Resistivity of germanium :
R = 1 / 2q * sqrt(Ng) * sqrt(u*h)
R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)
R = 1 /0.0001831
R = 5461.4964 ohm /cm
5461.4964 / 10000
0.546 ohm / μm
Answer:
C. Technician B
Explanation:
Excessive Galvanic activity:
To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.
Electrolysis problem:
When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.
In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.
Answer:
Explanation:
We are given that
d=1.9 cm=
Using 1m=100 cm
We have to find the electric field strength.
Using the formula
Mass of electron,m
Substitute the values
