Answer:
Fleming hand rule represents the direction of current in a generator's windings and induced current as a conductor is attached to a circuit such that it moves in a magnetic field.
Explanation:
Fleming hand rule represents the direction of current in a generator's windings and induced current as a conductor is attached to a circuit such that it moves in a magnetic field.
Fleming hand rule is used in the case of electric motors and electric generators.
Fleming hand rule is used to determine the following:
1. Direction of torque
2. Angular velocity
3. Angular acceleration
Answer:
a) 
b) 
Explanation:
From the question we are told that:
Cast-iron block Dimension:
Length
Width 
Feed
Depth 
Diameter 
Number of cutting teeth 
Rotation speed 
Generally the equation for Approach is mathematically given by
Therefore
Effective length is given as
a)
Generally the equation for Machine Time is mathematically given by
Where
Therefore
b)
Generally the equation for Material Removal Rate. is mathematically given by
Answer:
minimum electric power consumption of the fan motor is 0.1437 Btu/s
Explanation:
given data
area = 3 ft by 3 ft
air density = 0.075 lbm/ft³
to find out
minimum electric power consumption of the fan motor
solution
we know that energy balance equation that is express as
E in - E out = 
......................1
and at steady state = 0
so we can say from equation 1
E in = E out
so
minimum power required is
E in = W = m 
= 
put here value
E in =
E in = 
E in = 0.1437 Btu/s
minimum electric power consumption of the fan motor is 0.1437 Btu/s
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:
You can look it up
Explanation: if you don't know what it is look it up on .
