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Galina-37 [17]
2 years ago
8

49. A solenoid coil with a resistance of 30 ohms and an inductance of 200 milli-henrys, is connected to a 230VAC, 50Hz supply. D

etermine the solenoid impedance. *
49. A solenoid coil with a resistance of 30 ohms and an inductance of 200 milli-henrys, is connected to a 230VAC, 50Hz supply. Determine the solenoid impedance.
Engineering
1 answer:
Leya [2.2K]2 years ago
8 0

Answer:

69.59 ohms

Explanation:

Given that

L=200\ mH

F=50\ hZ

R=30\ ohms

For inductance

X_L=2\times \pi\times f \times LX_L\\=2\times \pi \times 50\times 200\times 10^{-3}\\=62.8\ ohm

R=30\ ohmsImpedance\\Z=\sqrt{R^2+X_L^2}\\Z=\sqrt{30^2+62.8^2}\ ohms\\Z =69.59\ ohms

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The u velocity component of a steady, two-dimensional, incompressible flow field is u = 3 ax 2 - 2 bxy, where a and b are consta
Kruka [31]

Answer:

The velocity component v is -6axy+2by^2+f(x)

Explanation:

Given that,

The velocity component of a steady, two-dimensional

u=3ax^2-2bxy

We need to calculate the function of x

Using given equation

u=3ax^2-2bxy

Where, a and b is constant

On differential

\dfrac{du}{dx}=6ax-2by

We need to calculate the velocity component v

Using equation of velocity

\dfrac{dv}{dy}=-\dfrac{du}{dx}-\dfrac{dw}{dz}

Put the value into the formula

\dfrac{dv}{dy}=-6ax+2by-0

Now, on integration w.r.t y

v=-6axy+2by^2+f(x)

Hence, The velocity component v is -6axy+2by^2+f(x)

4 0
2 years ago
The 2-Mg concrete pipe has a center of mass at point G. If it is suspended from cables AB and AC, determine the diameter of cabl
Lapatulllka [165]

Answer:

d_ab = 0.01189 m or 11.89 mm

Explanation:

Given:

- Weight of the block W = 2 Mg

- The diameter of cable AC, d_ac = 10 mm

Find:

- determine the diameter of cable AB so that the average normal stress in this cable is the same as in the cable AC.

Solution:

- We will apply equilibrium conditions on the given structure:

             Sum of forces in vertical y direction = 0

                        F_ab*cos(30) + F_ac*sin(45) - W = 0

   Where,         W = 2*10^3 * 9.81 = 19.62 KN

                       F_ab*cos(30) + F_ac*sin(45) - 19.62 = 0   .... 1

             Sum of forces in horizontal x direction = 0

                        F_ab*sin(30) - F_ac*cos(45) = 0               ..... 2

- Now solve Equation 1 and 2 simultaneously:

  From Eq 2:   F_ac*cos(45) = F_ab*sin(30)

  Input Eq 1:    F_ab * (cos(30) + sin(30)) = 19.62

                       F_ab = 19.62 /  (cos(30) + sin(30))

                       F_ab =  14.36 KN

  Input Eq 2:   F_ac = 14.36*sin(30) / cos(45)

                      F_ac = 10.16 KN

- Compute the cross-section areas of the both cables:

                       A_ac = pi*d_ac^2 / 4 = pi*(0.01)^2 / 4

                       A_ac = pi*d_ab^2 / 4 = pi*(d_ab)^2 / 4

- Compute the normal stress in both cables:

                       Q_ac = F_ac / A_ac

                       Q_ab = F_ab / A_ab

We know that: Q_ab = Q_ac

                        F_ac / A_ac = F_ab / A_ab

- Plug in the values:

                       F_ac / F_ab = A_ac / A_ab

                       10.16 / 14.36 = (pi*(0.01)^2 / 4) / (pi*(d_ab)^2 / 4)

                       d_ab = sqrt (14.355*0.01^2 / 10.16)

                       d_ab = sqrt (0.000141338)

                       d_ab = 0.01189 m or 11.89 mm

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2 years ago
Question 14<br> Which of these is true of a spray wash cabinet?
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Consider the following class definitions: class smart class superSmart: public smart { { public: public: void print() const; voi
arsen [322]

Answer:

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Explanation:

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2 years ago
A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the mome
Vikki [24]

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

                           mag(BC) = sqrt ( 48^2 + 36^2) = 60 in

                           BE / BC = 45 / 60 = 0.75

Hence,                E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in

- Find unit vector EF:

                           mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

                           vec(EF) = < -15 , -110 , 30 >

                           unit(EF) = (1/115) * < -15 , -110 , 30 >

- Tension            T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb

- Find unit vector AD:

                           mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

                           vec(AD) = < 48 , -12 , 36 >

                           unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >

                           unit (AD) = <0.7845 , -0.19612 , 0.58835 >

Next:

                           M_AD = unit(AD) . ( E x T_EF)

                           M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right]

                            M_AD = 1835.73 + 116.49528 - 593.0568

                            M_AD = 1359.17 lb-in

3 0
2 years ago
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