Answer:
0.5274 lagging.
Explanation:
So, we are given the following data or parameters or information which is what we need in order to solve this kind of question effectively. So, we have;
=> " For the given network:
ZR = 100 [Ω] , ZL = j124 [Ω], and ZC = - j89 [Ω]"
And we are asked to find the power factor. The power factor can be found or determined or calculated as;
7eq= 100 + j ( 24) (-89).
7eq = 124 × 89 - j89 × 100/ 100 + j35
7 eq= 133.81 L- 58.17.
Thus, the power factor is equal to;
=> Cos (58.17)= 0.5274 lagging.
Hence, the correct answer to this question is 0.5274 lagging.
Answer:
hello below is missing piece of the complete question
minimum size = 0.3 cm
answer : 0.247 N/mm2
Explanation:
Given data :
section span : 10.9 and 13.4 cm
minimum load applied evenly to the top of span : 13 N
maximum load for each member ; 4.5 N
lets take each member to be 4.2 cm
Determine the max value of P before truss fails
Taking average value of section span ≈ 12 cm
Given minimum load distributed evenly on top of section span = 13 N
we will calculate the value of by applying this formula
= = 1.56 * 10^-5
next we will consider section ; 4.2 cm * 0.3 cm
hence Z (section modulus ) = BD^2 / 6
= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8
Finally the max value of P( stress ) before the truss fails
= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )
= 0.247 N/mm2
Answer:
insert the disk in the laptop, wait of a minute or 2 and then a folder will open in my PC.
The equatorial diameter of the earth is 12,756 km.
Therefore the equatorial radius is
r = 12756/2 = 6378 km.
The angular velocity of the earth is
ω = (2π radians)/(24 hours) = 0.2618 rad/h
The tangential velocity of a person at the equator is
v = rω
= (6378 km)*( 0.2618 rad/h)
= 1670 km/h
Answer: 1670 km/h
I hope this helps you.
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Smile and have an amazing day ; )
Answer:
Can you please provide me with the diagram?
so I help. you out