As the length increases, resistance increases, as a result current decreases.
A. nucleus
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Answer:
3.6ft
Explanation:
Using= 2*π*sqrt(L/32)
To solve for L, first move 2*n over:
T/(2*π) = sqrt(L/32)
Next,eliminate the square root by squaring both sides
(T/(2*π))2 = L/32
or
T2/(4π2) = L/32
Lastly, multiply both sides by 32 to yield:
32T2/(4π2) = L
and simplify:
8T²/π²= L
Hence, L(T) = 8T²/π²
But T = 2.1
Pi= 3.14
8(2.1)²/3.14²
35.28/9.85
= 3.6feet
A. The proeutectoid phase is Fe₃c because 0.95 wt/c is greater than the eutectoid composition which is 0.76 wt/c
b. We determine how much total territe and cementite form, we apply the lever rule expressions yields.
Wx = (fe₃c-co/cfe₃ c-cx = 6.70- 0.95/6.70- 0.022 = 0.86
The total cementite
Wfe₃C = 10-Cx/ Cfe₃c -Cx = 0.95 - 0.022/6.70 - 0.022 = 0.14
The total cementite which is formed is
(0.14) × (3.5kg) = 0.49kg
c. We calculate the pearule and the procutectoid phase which cementite form the equation
Ci = 0.95 wt/c
Wp = 6.70 -ci/6.70 - 0.76 = 6.70 -0.95/6.70 - 0.76 = 0.97
0.97 corresponds to mass.
W fe₃ C¹ = Ci - 0.76/5.94 = 0.03
∴ It is equivalent to
(0.03) × (3.5) = 0.11kg of total of 3.5kg mass.
