Answer:
a. 1.91 b. -8.13 mm
Explanation:
Modulus =stress/strain; calculating stress =F/A, hence determine the strain
Poisson's ratio =(change in diameter/diameter)/strain
Answer:
a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4
b) 3.926 x 10∧-4, -2.626 x 10∧-4
c) 6.552 x 10∧-4, 6.5 x 10∧-5
Explanation:
a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4
b) 3.926 x 10∧-4, -2.626 x 10∧-4
c) 6.552 x 10∧-4, 6.5 x 10∧-5
The explanation is shown in the attachment. I hope i have been able to help.
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:
not sure if this helps but
