Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
Answer:
I always thought it was so that the older wire could not have a problem and have another electrician must come back and fix it.
Explanation:
Answer:
6.6 kilo volts = 6.6 k volts
Explanation:
A prefix is a word, number or a letter that is added before another word. In physics we have different prefixes for the exponential powers of 10, that are placed before units in place of those powers. Some examples are:
deci (d) ------ 10⁻¹
centi (c) ------ 10⁻²
milli (m) ------ 10⁻³
kilo (k) ------ 10³
mega (M) ----- 10⁶
giga (G) ------ 10⁹
We have:
6600 volts
converting to exponential form:
=> 6.6 x 10³ volts
Thus, we know that the prefix of kilo (k) is used for 10³.
Hence,
=> <u>6.6 kilo volts = 6.6 k volts</u>