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:
the value of conductivity in IP is 
Explanation:
Given that
Thermal conductivity K=14.54 W/m.K
This above given conductivity is in SI unit.
SI unit IP unit Conversion factor
m ft 0.3048
W Btu/hr 0.293
The unit of conductivity in IP is Btu./ft.hr.F.
Now convert into IP divided by 1.73 factor.

So


So the value of conductivity in IP is 
Answer:
the life (N) of the specimen is 46400 cycles
Explanation:
given data
ultimate strength Su = 1600 MPa
stress amplitude σa = 900 MPa
to find out
life (N) of the specimen
solution
we first calculate the endurance limit of specimen Se i.e
Se = 0.5× Su .............1
Se = 0.5 × 1600
Se = 800 Mpa
and we know
Se for steel is 700 Mpa for Su ≥ 1400 Mpa
so we take endurance limit Se is = 700 Mpa
and strength of friction f = 0.77 for 232 ksi
because for Se 0.5 Su at
cycle = (1600 × 0.145 ksi ) = 232
so here coefficient value (a) will be
a =
a =
a = 2168.3 Mpa
so
coefficient value (b) will be
a = -
log
b = -
log
b = -0.0818
so no of cycle N is
N =
put here value
N =
N = 46400
the life (N) of the specimen is 46400 cycles
Answer: Tech B
Explanation:
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