Answer:
All Brake lights are dimmer than normal because high resistance in the brake switch could be the cause according to Technician B.
Explanation:
According to Technician A
When the bulb is faulty then no current will flow through bulb and it will be open circuit.So no light will produce in bulb .
According to Technician B
When a high resistance inserted in series circuit the voltage across each resistance is reduced and this cause the light glow dimly.
Formula of resistance in series circuit
Rt=r1+r2+r3......
Answer:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV
Answer:
Recognize that there is a moral dilemma.
Determine the actor. ...
Gather the relevant facts. ...
Test for right versus wrong issues. ...
Test for right versus right paradigms. ...
Apply the resolution principles. ...
Investigate the trilemma options. ...
Make the decision.
Answer:
Wind energy is converted to Mechanical energy which is then converted in to electrical energy
Explanation:
In a wind mill the following energy conversions take place
a) Wind energy is converted into Mechanical energy (rotation of rotor blades)
b) Mechanical energy is converted into electrical energy (by using electric motor)
This electrical energy is then used for transmission through electric lines.
Answer:
± 0.003 ft
Explanation:
Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.
Let L' = our distance and L = our tape measure
So, L' = 100L
Now by error determination ΔL' = 100ΔL
Now ΔL' = ± 0.30 ft
ΔL = ΔL'/100
= ± 0.30 ft/100
= ± 0.003 ft
So, the maxim error per tape is ± 0.003 ft
