Answer:You are correct, no need to change.
Explanation:
Answer: 24 pA
Explanation:
As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.
Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵ Ω cm.
The resistance R of a given resistor, is expressed by the following formula:
R = ρ L / A
Replacing by the values for resistivity, L and A, we have
R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2
R = 2.1. 10¹¹ Ω
Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:
I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA
Answer: the standard deviation STD of machine B is s (Lb) = 0.4557
Explanation:
from the given data, machine A and machine B produce half of the rods
Lt = 0.5La + 0.5Lb
so
s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)
but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)
so we substitute
s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)
0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)
0.64 = 0.25 + s²(Lb) + 0.4s(Lb)
s²(Lb) + 0.4s(Lb) - 0.39 = 0
s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2
s (Lb) = 0.4557
therefore the standard deviation STD of machine B is s (Lb) = 0.4557
Answer:
Technician A only
Explanation:
Rapid wear of the sealing ring and the seal ring groove are result of excessive end play
Answer:
The time required to elute the two species is 53.3727 min
Explanation:
Given data:
tA = retention time of A=16.63 min
tB=retention time of B=17.63 min
WA=peak of A=1.11 min
WB=peak of B=1.21 min
The mathematical expression for the resolution is:
The mathematical expression for the time to elute the two species is:
Here
ReB = 1.5
