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tatuchka [14]
3 years ago
15

The direction of rotation of a dc series motor or a universal motor connected to a dc power source A) depends on the polarities

of the armature and field currents. B) depends exclusively on the polarity of the armature current. C) depends exclusively on the polarity of the field current. D) depends on the connection of the compensating winding.
Engineering
1 answer:
Solnce55 [7]3 years ago
8 0

Answer:

<em>A) depends on the polarities of the armature and field currents.</em>

Explanation:

In the DC series motor, or a universal motor, the field and armature winding are connected in series. This means that if the polarity of the input DC voltage is changed, the direction of the field & armature will also change, resulting in the change of direction of rotation of the motor. This means that the direction of rotation of a DC series motor, or a universal motor connected to a DC power source can be reversed by changing the polarity of the field current, or the armature current.

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Concerned with the number of maintenance visits the rocket can undergo before being out of service, you have been informed that
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Answer:

(a) Mn = M₁ + (n-1) (M₂ -M₁) = 1 + (n- 1) 1 = n (b) n > 10 (exceed 10) or n =11 (c) n >50 or n= 51

After making a journey of 51 times, the rocket will be discarded

Explanation:

Solution

(a) Let Mn denotes the number of  maintenance visits after the nth journey

Then M₁ = 1 , M₂ = 1 +M₁ = 2, M₃ = 1 +M₂ = 3

We therefore, notice that M follows an arithmetic sequence

So,

Mn = M₁ + (n-1) (M₂ -M₁)

= 1 + (n- 1) 1 = n

or Mn =n

(b)  For what value of n we will get  fro Mn > 10

Thus,

n > 10 (exceed 10) or n =11

(c)Similarly of Mn is greater than 50 or Mn>50, the rocket will not be used or reused

So,

n >50 or n= 51

After making a journey of 51 times, the rocket will be discarded

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Answer:

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Except the Table of Contents
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A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i
dlinn [17]

Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

slope in xz plane is

tan\theta = \frac{a_x}{g +a_z}

tan\theta = \frac{4}{9.81+0}

tan\theta =0.4077

the maximum height of water surface at mid of inclination is

\Delta h = \frac{d}{2} tan\theta

            =\frac{0.4}{2}0.4077

\Delta h  0.082 cm

the maximu height of wwater to avoid spilling is

h_{max} = h_{tank} -\Delta h

            = 60 - 8.2

h_{max} = 51.8 cm

the height requird if no spill water is h_{max} = 51.8 cm

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