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:
Fracture.
Explanation:
Fracture describes how a mineral looks when it breaks apart in an irregular way.
Answer:
B A and C
Explanation:
Given:
Specimen σ
σ
A +450 -150
B +300 -300
C +500 -200
Solution:
Compute the mean stress
σ
= (σ + σ)/2
σ
= (450 + (-150)) / 2
= (450 - 150) / 2
= 300/2
σ = 150 MPa
σ
= (300 + (-300))/2
= (300 - 300) / 2
= 0/2
σ = 0 MPa
σ
= (500 + (-200))/2
= (500 - 200) / 2
= 300/2
σ = 150 MPa
Compute stress amplitude:
σ
= (σ - σ)/2
σ
= (450 - (-150)) / 2
= (450 + 150) / 2
= 600/2
σ = 300 MPa
σ
= (300- (-300)) / 2
= (300 + 300) / 2
= 600/2
σ = 300 MPa
σ
= (500 - (-200))/2
= (500 + 200) / 2
= 700 / 2
σ = 350 MPa
From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.
Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.
In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.
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