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Sunny_sXe [5.5K]
3 years ago
9

PLEASE HELP!

Physics
1 answer:
OlgaM077 [116]3 years ago
4 0
1. a-ask a question
2.c-the results
3.c-the question we pose
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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
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