Answer: C. Voltage
Explanation:
Here are some other words as well.
potential, voltage, potential drop, potential difference.
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Answer:
a. Solid length Ls = 2.6 in
b. Force necessary for deflection Fs = 67.2Ibf
Factor of safety FOS = 2.04
Explanation:
Given details
Oil-tempered wire,
d = 0.2 in,
D = 2 in,
n = 12 coils,
Lo = 5 in
(a) Find the solid length
Ls = d (n + 1)
= 0.2(12 + 1) = 2.6 in Ans
(b) Find the force necessary to deflect the spring to its solid length.
N = n - 2 = 12 - 2 = 10 coils
Take G = 11.2 Mpsi
K = (d^4*G)/(8D^3N)
K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in
Fs = k*Ys = k (Lo - Ls )
= 28(5 - 2.6) = 67.2 lbf Ans.
c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.
For C = D/d = 2/0.2 = 10
Kb = (4C + 2)/(4C - 3)
= (4*10 + 2)/(4*10 - 3) = 1.135
Tau ts = Kb {(8FD)/(Πd^3)}
= 1.135 {(8*67.2*2)/(Π*2^3)}
= 48.56 * 10^6 psi
Let m = 0.187,
A = 147 kpsi.inm^3
Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi
Ssy = 0.50 Sut
= 0.50(198.6) = 99.3 kpsi
FOS = Ssy/ts
= 99.3/48.56 = 2.04 Ans.
Answer:
the MTTF of the transceiver is 50.17
Explanation:
Given the data in the question;
failure modes = 0.1 failure per hour
system reliability = 0.85
mission time = 5 hours
Now, we know that the reliability equation for this situation is;
R(t) = [ 1 - ( 1 - 
)³] 
so we substitute
R(5) = [ 1 - ( 1 - 
)³] 
= 0.85
[ 1 - ( 1 - 
)³] = 0.85
[ 1 - ( 0.393469 )³] = 0.85
[ 1 - 0.06091 ] = 0.85
0.9391 = 0.85
= 0.85 / 0.9391
= 0.90512
MTTF = 5 / -ln( 0.90512 )
MTTF = 50.17
Therefore, the MTTF of the transceiver is 50.17
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