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A motor with a brush-and-commutator arrangement has a circular coil with radius 2.5cm and 150 turns of wire. The magnetic field

Amanda [17]

Answer:

(a) 0.81 V

(b) 0.52 V

Explanation:

Number of turns, N = 150

Radius, r = 2.5 cm = 0.025 m

Magnetic field, B = 0.060 T

f = 440 rev/min = 440 / 60 = 7.33 rps

A.

The maximum emf is given by

e = N x B x A x 2 x π x f

e = 150 x 0.060 x 3.14 x 0.025 x 0.025 x 2 x 3.14 x 7.33

e = 0.81 V

B.

The back emf is given by

e' = 2e / π = 2 x 0.81 / 3.14 = 0.52 V

3 0

3 years ago

Does a North magnet and south magnet repel or attract?

NNADVOKAT [17]

Answer:

A north magnet attracts a south magnet

Explanation:

The opposite polar magnetic field will attract each other while the same polar magnetic fields will repel each other.

5 0

3 years ago

Read 2 more answers

What are the two forces acting on the upper magnet?

solniwko [45]

Answer:

b

Explanation:

I'm not so sure but it makes sense

8 0

3 years ago

in three situations, a briefly applied horizontal force changes the velocity of a hockey puck that slides over frictionless ice.

victus00 [196]

The work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

According to Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

F = ma

$F= \frac{mv}{t}$

The force applied to an object increases with increases in the velocity of the object.

In the given diagram, the resultant velocity of the puck is calculated as follows;

Figure a:

$\Delta v = v_f -v_i\\\\\Delta v = 5 - 6 = - 1 \ m/s$

Figure b:

$v = \sqrt{4^2 + 3^2} \\\\v = 5 \ m/s$

Figure c:

$\Delta v = 4 - (-2)\\\\\Delta v = 6 \ m/s$

Thus, the work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

Learn more here:brainly.com/question/19498865

4 0

3 years ago

Air enters a turbine operating at steady state at 8 bar, 1400 K and expands to 0.8 bar. The turbine is well insulated, and kinet

vladimir2022 [97]

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

Where

$T_1 =$ Temperature at inlet of turbine

$T_2 =$ Temperature at exit of turbine

$P_1 =$ Pressure at exit of turbine

$P_2 =$ Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

$m_i = m_0 = m$

$m(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W$

Where,

m = mass

$m_i$ = mass at inlet

$m_0$ = Mass at outlet

$h_i$ = Enthalpy at inlet

$h_0$ = Enthalpy at outlet

W = Work done

Q = Heat transferred

$V_i$ = Velocity at inlet

$V_0$ = Velocity at outlet

$Z_i$ = Height at inlet

$Z_0$ = Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + W$

$W = h_i -h_0$

Using the relation T-P we can find the final temperature:

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

$\frac{T_2}{1400K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}$

$T_2 = 725.126K$

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

So:

$W = h_i -h_0$

$W = C_p (T_1-T_2)$

$W = 1.005(1400-725.126)$

$W = 678.248kJ/Kg$

Therefore the maximum theoretical work that could be developed by the turbine is 678.248kJ/kg

5 0

4 years ago