Explanation:
first changing kilo ohm to ohm
860000 = 860 kΩ
and change 34 micro ampare to ampare
34 μA=3.4×10^-5
recalling the equation V=I*R
V= 3.4×10^-5×860000
v=29.24
Answer:
it is not possible to place the wires in the condui
Explanation:
given data
total area = 2.04 square inches
wires total area = 0.93 square inches
maximum fill conduit = 40%
to find out
Can it is possible place wire in conduit conduit
solution
we know maximum fill is 40%
so here first we get total area of conduit that will be
total area of conduit = 40% × 2.04
total area of conduit = 0.816 square inches
but this area is less than required area of wire that is 0.93 square inches
so we can say it is not possible to place the wires in the conduit
Answer:
attached below is the detailed solution and answers
Explanation:
Attached below is the detailed solution
C(iii) : versus the parameter C
The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C
Answer:
If you mean two sides are 7 and two sides are 14 then you'd have 42
and for the second you'd have 14
Explanation:
7 + 7 = 14, 14 + 14 = 28, 14 + 28 = 42
3 + 3 = 6, 4 + 4 = 8, 8 + 6 = 14