The current will lag the voltage in AC circuit that contains both resistance and inductance.
Answer: C
Explanation
There is no inductance only circuits in reality.
The circuits containing inductance has also a lower amount of resistance.
The current flows in both resistance and inductance.
There is a drop in the total voltage in resistance and inductance giving rise to the voltage applied in the coil when connected in a series.
An example being inductance coil an AC circuit connected to both resistance and inductance in series.
From the vector diagram, this conclusion can be drawn.
The answer is the test is being tested towards the lungs the test is done by scanning your body the tools are called “the x rat visional lock space” and the rubber tool is called a deeldo it’s purple with a pencil looking shape perfect for the body.
Answer: resistor
Explanation: Not quite sure. Need more research
Answer:
a) the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb
Explanation:
Given the data in the question;
Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in
Diameter d = 3/16 in
Unit weight w = 490 pcf
First we determine the area of the wire;
A = π/4 × d²
we substitute
A = π/4 × (3/16)²
A = 0.0276 in²
Next we get the Volume
V = Area × Length of wire
we substitute
V = 0.0276 × 4800
V = 132.48 in³
Weight of the steel wire will be;
W = Unit weight × Volume
we substitute
W = 490 × ( 132.48 / 12³ )
W = 490 × 0.076666
W = 37.57 lb
a) the maximum tensile stress due to the weight of the wire;
σ
= W / A
we substitute
σ
= 37.57 / 0.0276
= 1361.23 psi
Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi
Maximum load P that the wire can safely support its lower end will be;
P = ( σ
- σ
)A
we substitute
P = ( 24000 - 1361.23 )0.0276
P = 22638.77 × 0.0276
P = 624.83 lb
Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb