Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?
a. 2.17000000 A-2.258571429 A
b. 2.20000000 A-2.29000000 A
c. 2.211445 A-2.30000000 A
d. 2.20144927 A-2.29130435 A
e. 2.00000000 A-2.30000000 A

Answer:

Bluray

DVD

CD

Explanation:

Blu ray can hold 25gb per layer

Dvd can hold 4.7GB on a single layer

Cd can hold around 737 mb

Also, dvds can go up to 2 layers

Blu ray can go up to 4

Answer:

CD

 ↓

DVD

  ↓

Blu-ray

Explanation:

Answer:

staples

Door or window

Explanation:

staples.- can hit the internal core and risk of electrocution.

Door or window - can damage the outer core

staples.- can hit the internal core and risk of electrocution or Door or window - can damage the outer core.

A length of flexible electrical power cable (flex) with a plug on one end and one or more sockets on the other end (often of the same kind as the plug) is known as an extension cord (US), power extender, drop cord, or extension lead (UK).

The phrase is also used to describe extensions for various types of cabling, but it mainly refers to mains (home AC) extensions.

The phrase "adapter cord" may be used if the plug and power outlet are of different types. Although they can be produced up to 300 feet long, most extension cords are between two and thirty feet long.

Therefore, staples.- can hit the internal core and risk of electrocution or Door or window - can damage the outer core.

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Answer:

The bronze age

Answer: New materials

Explanation: I read it in the article.

Answer:  1804.61 lbs

Explanation:

Given, diameter of concrete column = 390 mm

= 0.390 m  [ 1 mm =0.001 m]

⇒ Radius = [tex]\dfrac{0.390}{2}=0.195\ m[/tex]

also, Length = 2.8 m

Volume of column  (cylindrical)= [tex]\pi r^2l[/tex]

[tex]=(3.14)(0.195)^2(2.8)[/tex]

[tex]=0.3343158\ m^3[/tex]

Density  [tex]=\dfrac{Mass }{Volume}=2.45\ Mg/m^3[/tex]  

⇒ Mass = 2.45 x (Volume)

= 2.45 (0.3343158) Mg

= 0.81907371 Mg

⇒ Mass = 819.07271 kg  [1 Mg =1000 kg]

Weight = [tex]{mass}\times{gravity}={819.07271}\times{9.8}=8026.91\ N[/tex]

1 N= 0.22482 lb

Then, Weight = [tex]0.22482\times 8026.91=1804.61\ lb[/tex]

Weight of the column in pounds.= 1804.61 lbs (approx)

Weight of the column is 1804.61 lbs.

Given that,

According to the given data, calculation are as follows,

Diameter = 0.39m

So, radius (r) = 0.39 [tex]\div[/tex] 2 = 0.195m

Volume of column = [tex]\pi r^{2}l [/tex]

By putting the value, we get

Volume = [ (3.14)[tex]\times[/tex] [tex](0.195)^{2} [/tex] [tex]\times[/tex] 2.8m ]

= [tex]0.3343158m^{3} [/tex]

Now, as we know that,

Density = Mass [tex]\div[/tex] Volume

Or, Mass (M) = Density [tex]\times[/tex] Volume

= 2.45 [tex]\times[/tex] 0.3343158

= 819.07271 kg

Now Weight = (M [tex]\times[/tex] Gravity )         Value of gravity is 9.8.

Weight = 819.07271 kg [tex]\times[/tex] 9.8

= 8026.91N

Where, 1N = 0.22482 lbs

So, Weight of the column = 1804.61 lbs

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Answer:

The amount of phase shift between input and output signal is important when measuring a common emitter amplifier circuit​.

Explanation:

the amount of phase shift between input and output signal is important when measuring a common emitter amplifier circuit​

In signal processing, phase distortion is change in the shape of the waveform, that occurs when the phase shift introduced by a circuit is not directly proportional to frequency.

In a common emitter amplifier circuit​ there is an 180-degree phase shift between the input and output waveforms.

Answer:

(a) km/s

(b) mm

(c) Gs/kg

(d) mmN

Explanation:

(a) Given:

m / ms => the proper SI unit should be m/s

So let's convert to m/s

m / ms = [tex]\frac{m}{ms}[/tex]

The denominator is ms which means milliseconds.

But milli = 10⁻³

∴ [tex]\frac{m}{ms}[/tex] = [tex]\frac{m}{10^{-3}s}[/tex] = [tex]\frac{10^3m}{s}[/tex]

Also 10³ = kilo (k)

[tex]\frac{10^3m}{s}[/tex] = [tex]\frac{km}{s}[/tex]

∴ m / ms = [tex]\frac{km}{s}[/tex] =  km/s

(b) Given:

μkm => the proper SI unit should be m

So let's convert to m

μkm = 10⁻⁶ km [since μ = 10⁻⁶]

=> μkm = 10⁻⁶ x 10³ m [since k = 10³]

=> μkm = 10⁻³ m

But 10⁻³ = milli (m)

=> μkm = 10⁻³ m = mm

∴ μkm = mm

(c) Given:

ks/mg => the proper SI unit should be s/kg

So let's convert to s/kg

ks/mg = 10³s / mg [since k = 10³]

=> ks/mg = 10³s / 10⁻³g [since m = 10⁻³]

=> ks/mg = 10³ x 10³ s / g

=> ks/mg = 10⁶ s / g

=> ks/mg = [tex]\frac{10^6 s}{g}[/tex]

Multiply the numerator and denominator by 10³

=> ks/mg = [tex]\frac{10^6 X 10^3s}{10^3 X g}[/tex]

=> ks/mg = [tex]\frac{10^9s}{10^3 X g}[/tex]

=> ks/mg = [tex]\frac{Gs}{10^3 X g}[/tex]  [since 10⁹ = Giga (G) ]

=> ks/mg = [tex]\frac{Gs}{kg}[/tex]  [since 10³ = Kilo (k) ]

∴ ks/mg = [tex]\frac{Gs}{kg}[/tex]  = Gs/kg

(d) Given:

km.μN => the proper SI unit should be mN or Nm

So let's convert to km.μN

km.μN = 10³m.μN                 [since k = 10³]

=> km.μN = 10³m.10⁻⁶N        [since μ = 10⁻⁶]

=> km.μN = 10⁻³ mN

=> km.μN = mmN             [since m = milli = 10⁻³]

∴ km.μN = mmN

John Smeatom, U.K. 18th century, was the first self-proclaimed, civil engineer in the 18th century and IS considered “the father of modern, civil engineering”.

hoped this helped! :)

Answer:

Sir Mokshagundam Visvesvaraya

Answer:

A) 0.0614 inches

b) The standard steel paper clip should float on water

Explanation:

The maximum diameter that the rod can have before it will sink

we can calculate this using this formula :

D = [tex](\frac{8\alpha }{\pi y } )^{\frac{1}{2} }[/tex] ----- 1

∝ = value of surface tension of water at 60⁰f  = 5.03×10^−3  lb/ft

y = 490 Ib/ft^3

input the given values into equation 1 above

D = [tex](\frac{8*(5.3*10^{-3}) }{\pi *490 } )^{\frac{1}{2} }[/tex]

   = 5.11 * 10^-3 ft   convert to inches

   = 5.11 *10^-3 ( 12 in/ 1 ft ) = 0.0614 inches

B) The diameter of a standard paper Cliphas = 0.036 inches

and the diameter of the rod = 0.0614. Hence the standard steel paper clip should float on water

Answer: The Government.

Explanation: Highways are usually a roads where people travel on. They are usually owned mainly by the government some are owned by private entities. Of the above options the most possible employer of Trevor is the government as majority of the highways are usually owned by the government.

The government provide several services on the highways such as roads maintenance, security, accidents and emergency cares etc

Answer: I think it's C, I hope this helps ;)

Answer:

Kindly check explanation

Explanation:

(a) μMN, (b) N/μm, (c) MN/ks2, and (d) kN/ms.

(a) μMN = (10^-6 * 10^6) = 10^(-6 + 6) = 10^0 = N

b) N/μm = N / 10^-6 m = 10^6 * N/m = MN/m

(c) MN/ks2 = 10^6N / (10^3 s)^2

10^6 N / 10^6s^2 = 10^6 * 10^-6 N /s^2 = N/s^2

D) kN/ms = 10^3N / 10^-3 s = 10^3 * 10^3 * N/s = 10^6N / s = MN/s

2)

a) 0.000431kg = 431 × 10^6 kg = 431 * 10^9g = 431Gg

b) 35.3 × 10^3 N

10^3 = kilo(K)

35.3 KN

C) 0.00532km = 5.32 * 10^3 km = 5.32 * 10^3 * 10^3 = 5.32 * 10^6 = 5.32Mm

3) Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b) N/nm, and (c) mN/(kg⋅μs).

a) Mg/ms = 10^6g / 10^-3s = 10^6 * 10^3 g/s = 10^9 g/s = Gg/s

b) N/nm = N / 10^-9 m = 10^9 N/m = GN/m

c) mN/(kg⋅μs) = 10^6N / kg(10^-6s) = 10^12N/(kg.s)

= TN/(kg.s)

Answer:

B) 5.05

Explanation:

The wall thickness of a pipe is the difference between the diameter of outer wall and the diameter of inner wall divided by 2. It is given by:

Thickness of pipe = (Outer wall diameter - Inner wall diameter) / 2

Given that:

Inner diameter = ID = 25 ± 0.05, Outer diameter = OD = 35 ± 0.05

Maximum outer diameter = 35 + 0.05 = 35.05

Minimum inner diameter = 25 - 0.05 = 24.95

Thickness of pipe = (maximum outer wall diameter - minimum inner wall diameter) / 2 = (35.05 - 24.95) / 2 = 5.05

or

Thickness = (35 - 25) / 2 + 0.05 = 10/2 + 0.05 = 5 + 0.05 = 5.05

Therefore the LMC wall thickness is 5.05

Answer:

Qsp > ksp

Explanation:

given data

Ag2CrO4 = 9.0 × [tex]10^{-12}[/tex]

solution

new [[tex]Ag^+[/tex]] = [tex]\frac{0.005 \times 200}{200+300}[/tex] = 0.002 M     .................1

and

new [[tex]CrO^{2-}_4[/tex]] = [tex]\frac{0.002 \times 300}{200+300}[/tex]  = 0.0012M    ..............2

so here solubility equation will be as

[tex]Ag_2CrO_4(s)[/tex] ⇄  2[tex]Ag^+[/tex] (aq) + [tex]CrO^{2-}_4[/tex] (aq)    ......................3

so

Qsp = [tex][Ag^+]^2[/tex] [ [tex]CrO^{2-}_4[/tex]]

put here value

Qsp = (0.002)² × (0.0012)

Qsp = 4.80 × [tex]10^{-9}[/tex]  

so that we can say that Qsp > ksp

Answer:

the elevation of the water surface over the obstruction is highest at 1.341 m

the maximum height of the obstruction h = 0.159 m

Explanation:

From the given information:

The diagrammatic expression for the water profile showing a rectangular channel  with 3 cm width carrying a 4 m3/s of water at a depth of 1.5 m with an obstruction 15 cm high is placed across the channel can be seen in the diagram attached below.

To calculate the elevation of the water surface over the obstruction, we need to  determine the following:

a. the velocity of the channel

b. the froude number at the upstream of the obstruction

c. the specific energy level

To start with the velocity V of the channel.

[tex]V_1 = \dfrac{Q}{A} \\ \\ V_1 = \dfrac{4 \ m^3/s}{(3 \times 1.5 ) m^2 } \\ \\ V_1 = \dfrac{4 \ m^3/s}{4.5 \ m^2}[/tex]

[tex]V_1 = 0.88 \ m/sec[/tex]

The froude number  at the upstream of the obstruction

[tex]F _{\zeta} = \dfrac{V_1}{\sqrt{gy__1}}[/tex]

[tex]F _{\zeta} = \dfrac{0.88}{\sqrt{9.81 \times 1.5}}[/tex]

[tex]F _{\zeta} = \dfrac{0.88}{\sqrt{14.715}}[/tex]

[tex]F _{\zeta} = \dfrac{0.88}{3.836}[/tex]

[tex]F _{\zeta} =0.229[/tex]

[tex]F _{\zeta} \simeq0.3[/tex]  which is less than the subcritical flow.

Similarly, the specific energy level for this process can be expressed as:

[tex]E_1 = \dfrac{V_1^2}{2g}+y_1[/tex]

[tex]E_1 = \dfrac{0.88^2}{2 \times 9.81 }+1.5[/tex]

[tex]E_1 = \dfrac{0.7744}{19.62 }+1.5[/tex]

[tex]E_1 =0.039469+1.5[/tex]

[tex]E_1 =1.539469[/tex]

[tex]E_1 \simeq1.54 \ m[/tex]

[tex]E_2 + \Delta z = E_1[/tex]

[tex]E_2= E_1 - \Delta z[/tex]

[tex]E_2= (1.54 - 0.15) \ m[/tex]

[tex]E_2=1.39 \ m[/tex]

Suppose ;

[tex]V_1 y_1 = V_2y_2\\ \\ Then; \ making \ V_2 \ the \ subject \ of \ the \formula\ we \ have: \\ \\ \\V_2 = \dfrac{V_1 y_1}{y_2} ---- (1)[/tex]

From the energy equation:

[tex]E_2 + \Delta z = E_1[/tex]

we can now substitute the above derived parameter and have :

[tex]E_2 + \Delta z=y_1 + \dfrac{V_1^2}{2g}[/tex]

[tex]E_2 =y_1 + \dfrac{V_1^2}{2g} - \Delta z[/tex]

[tex]E_2 =y_2 + \dfrac{V_2^2}{2g}[/tex]

replace the value of [tex]V_2[/tex] =[tex]\dfrac{V_1 y_1}{y_2}[/tex]  in equation (1), we have:

[tex]E_2 =y_2 + \dfrac{( \dfrac{V_1y_1}{y_2})^2}{2g}[/tex]

[tex]E_2 =y_2 + \dfrac{ {V_1^2y_1^2}}{2gy_2^2}[/tex]

[tex]E_2 \times y_2^2 =y^3_2 + \dfrac{ {V_1^2y_1^2}}{2g}[/tex]

[tex]y^3_2 - E_2y^2_2 + \dfrac{V_1^2y_1^2}{2g}=0[/tex]

Replacing our values now; we have:

[tex]y^3_2 - 1.39 \times y^2_2 + \dfrac{0.88^2 \times 1.5^2}{2 \times 9.81}=0[/tex]

[tex]y^3_2 - 1.39 \times y^2_2 + \dfrac{1.7424}{19.62}=0[/tex]

[tex]y^3_2 - 1.39 \times y^2_2 + 0.0888=0[/tex]

[tex]y^3_2 - 1.39 y^2_2 + 0.0888=0[/tex]

[tex]y_2 = 1.341 \ m \\ \\ y_2 = - 0.232 \ m \\ \\ y_2 = 0.281 \ m[/tex]

Therefore,the elevation of the water surface over the obstruction is highest at 1.341 m

What is the maximum height of the obstruction that will not cause a rise in the water surface upstream

In order to determine the maximum height , we need to first estimate the rise in water level surface of [tex]\Delta z[/tex]

[tex]\Delta z[/tex] = 1.5 - (1.341+ 0.15) m

[tex]\Delta z[/tex] =  (1.5 - 1.491) m

[tex]\Delta z[/tex] = 0.009 m

Finally, the maximum height of the obstruction h = (0.009 + 0.15 )m

the maximum height of the obstruction h = 0.159 m

Answer:

a) the spring rate is 3.333 N/mm

b) the minimum hole diameter for the compression spring is 44 mm

c) the total number of coils needed is 11.6

d) the solid length is 50.4 mm

Explanation:

a)

to calculate the mean spring coil diameter, we take a look at the expression from the relation;

D = Cd

where C is the spring index ( 10 ) and d is the diameter of helical compression spring (4 mm)

so we substitute

D = 10 × 4 = 40 mm

Torsional stiffness G for the tempered wire with diameter 4 mm is 77.2 Gpa ( 77.2 × 10³ Mpa)  ( obtained from Table: Mechanical properties of spring wires).

so when the spring is compressed, the spring force is given by the following expression(realtion)

Fs = k × ys

where ys is the deflection of the spring (15 mm) and k is the spring rate, Fs is the force (50N)

so we substitute

50N = k × 15mm

k = 50N / 15mm

k = 3.333 N/mm

∴  the spring rate is 3.333 N/mm

b)

to calculate the minimum hole diameter for the compression spring

Now the entire spring is within a hole in the  ground, therefore the hole should have a diameter equal to the outer diameter of the spring.

so D₀ = D + d

and from our initial equations, the mean spring coil diameter D = 40mm and the diameter of the helical compression spring d = 4mm

we substitute

D₀ = 40 + 4

D₀ = 44 mm

the minimum hole diameter for the compression spring is 44 mm

c)

Consider the following relation to calculate the total number of coils needed

Na coils are actually working to support the springs structure and its all dependent on the cut at the edge (end). ( from the table, Nt elates to Na)

Na = (d⁴G) / 8D³k

where the mean spring coil diameter D = 40mm and the diameter of the helical compression spring d = 4mm, G is the torsional stiffness  (77.2 × 10³ Mpa), the the spring rate k is 3.333 N/mm

so we substitute

Na = (4⁴(77.2 × 10³)) / ( 8(40³)(3.333))

Na = 19,763,200 / 1,706,496

Na = 11.6

the total number of coils needed is 11.6

d)

As the number of active coils and total number of coils are the same, we get the following relation;

Na = Nt

Nt which is also total number of coils

Now to calculate the solid length

Ls = d ( Nt + 1 )

so we substitute

Ls = 4 ( 11.6 + 1 )

Ls = 50.4 mm

the solid length is 50.4 mm

The spring rate is of the spring is 3.33N/mm, the minimum hole diameter required for the spring to operate in is 44mm while the number of coil needed is 11.6 or approximately 12 and the solid length of the spring is 50.4mm

Given data:

a) Spring rate:

[tex]k = f/[/tex]δ = [tex]50/15=3.33N/mm[/tex]

b) Minimum hole diameter;

The minimum hole diameter(D[tex]_i[/tex]) is the sum of diameter of the wire + D

[tex]D_i=40+4=44mm[/tex]

c) Total number of coils needed:

To solve this, we need to use a constant known as the modulus of rigidity and it's given as G = [tex]77.2*10^3N/mm[/tex]

From δ=[tex]\frac{8fD^3N}{Gd^4}[/tex]

Making N the subject, we would have

[tex]N=\frac{15*77.2*10^3*4^4}{8*50*40^3}=11.58[/tex]≈11.6 coils

The number of coils needed is 11.6 coils or approximately 12 coils.

d) The solid length;

The solid length formula is given as

[tex]L_s=(N+1)d=(11.6+1)*4=50.4mm[/tex]

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Answer:

see below

Explanation:

10 per hour

example: 37.5 hours in a week

10 x 37.5 hours per week x 52 weeks in a year

19,500 per year.

you can use the above calculation depending on how much you want to work per week.

Answer:

A. 0.66 Amps

Explanation:

Using ohms law, we can say that Voltage is equivalent to Current times Resistance.  We are given the voltage and the resistance of the circuit, so we simply need to find the current.

V = IR

Solve for I, where V = 2volts and R = 3ohms.

V = IR

V * 1/R = I * R * 1/R

I = V/R

I = 2/3 Amps

Hence, we should choose option A, 0.66 Amps for the current in this simple circuit.

Cheers.

Answer:

The transistor goes into cutoff ( A )

Explanation:

In a voltage-divider biased npn transistor, if the upper voltage-divider resistor (the one connected to VCC) opens the transistor goes into cutoff

Voltage-divider biased npn transistor is a transistor that consists of some resistors and this resistors help to divide and distribute the voltage entering the transistor into its correct levels ( this is the most popular means of biasing a BJT transistor )

Answer:

D. All the above

Explanation:

A is correct. The theory of formation of concrete is based on this phenomenon of formation of voids because, a coarse aggregate aggregate on its own, contain voids. When a fine aggregate like sand is added, it occupies these voids. Further, when finely powdered cement is added to the mixture, it further occupies the voids between the sand particles.

B is correct. The bulking of sand should be taken into account when volumetric proportioning of the aggregates is adopted. Otherwise, less quantity of concrete per bag of cement will be produced, which naturally will increase the cost of concrete. This is because sand particles when left on their own, do not pack properly near each other due to sand buckling, leaving too many void spaces between the sand particles.

C is correct . The dry sand and the sand completely flooded with water have practically the same volume. This is due to the bulking of sand. When water is added to the sand, up to complete flooding, the sand particles pack near each other and the amount of bulking of sand is decreased. The addition of the volume of water, leads to a decrease in the volume of the sand due to the extra voids. These cancels out to practically leave the dry sand and the flooded sand with the same volume.

The names of each of the crystal structures for each of the given metal alloys are; Alloy A is BCC structure; Alloy B is SC structure; Alloy C is BCC structure

To solve for each of the three given alloys, we will need to do so by trial and error method to calculate the density and compare it to the value given in the problem.

The density formula is;

ρ = nA/(V_c × N).

Where;

n is 1, 2 and 4 for SC, BCC & FCC crystal structures respectively.

A is atomic weight

V_c is volume expressed as a³.

a is given by 2R,4R/√3, 2R√2 for SC, BCC & FCC crystal structures respectively with R being the atomic radius.

N is avogadros number of atoms and has a constant value of 6.022 × 10^(23) atoms/mol

A = 43.1 g/mol

If we assume it's a BCC structure, then;

n = 2

V_c = a³ = (4R/√3)³ = (4 × 1.22 × 10^(-8) × 1/√3)³ = 22.4756 × 10^(-24) cm³

N = 6.022 × 10^(23) atoms/mol

Plugging in the relevant values into the density formula, we have;

ρ = (2 × 43.1)/(22.4756 × 10^(-24) × 6.022 × 10^(23)) ≈ 6.4 g/cm³

This value is same as the one in the question, thus metal alloy A is a BCC crystal structure

A = 184.4 g/mol

If we assume it's a SC structure, then;

n = 1

Thus;

V_c = a³ = (2 × 1.46 × 10^(-8))³ = 24.897 × 10^(-24) cm³

Plugging in the relevant values into the density formula, we have;

ρ = (1 × 184.4)/(24.897 × 10^(-24) × 6.022 × 10^(23)) ≈ 12.3 g/cm³

This value is same as the one in the question, thus metal alloy B is an SC crystal structure

A = 91.6 g/mol

If we assume it's a BCC structure, then;

n = 2

V_c = a³ = (4R/√3)³ = (4 × 1.37 × 10^(-8) × 1/√3)³ = 31.67 × 10^(-24) cm³

Plugging in the relevant values into the density formula, we have;

ρ = (2 × 91.6)/(31.67 × 10^(-24) × 6.022 × 10^(23)) ≈ 9.6 g/cm³

This value is same as the one in the question, thus metal alloy C is an SC crystal structure.

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Answer:

A) magnitude of the line current = 15.24 A

B) 6583.94 v

Explanation:

Given data :

magnitude of line voltage at terminal = 6600 v

load impedance = 240 - j70 Ω/∅

impedance of load fed from a line = 0.5 + j4 Ω/∅

attached below is the detailed solution

Answer:

First, there are 4 atoms in the FCC unit cell.

The unit volume of the unit can be calculated based on the atomic radii, in which case the hypotenuse of the cube of the unit cell would be 4 X 0.1278  

Next, find the weight of one atom by taking (63.55g / mol) X (mols / 6.0221415 x 10 ^ 23 atoms)

So now you have all the numbers you need. Take the weight of the atom X 4 divided by the cube volume taken from the hypotenuse.

Explanation:

Answer:

B. with each component having an individual address

Explanation:

A data bus is a system within a computing system, that consists of a  set of wires or connectors, that provides transportation for data. A data bus can transfer data and information through a computing system, or between two or more computing systems. Each component in the computer has its own unique address by which data or information is sent to it, to or fro the central processing unit, and the data or information travelling through the data bus, which serves as the highway, reaches each component using this address. This is analogous to the postal service system of moving letters and packages, through a series of networks that locates the address of the receiver.

Answer:

Is

Explanation:

Answer:

1) Visual Communication involves the use of images ans symbolism for the exchange of ideas from a source to the intended audience.

Visual Communication can help enhance the message conveyed by the speaker when the message being conveyed is new to the audience, such as instructional or educational presentation by enable the audience to more quickly understand the the point of view of the speaker

2) A particular situation where it is best to use visual communication is in a communication that involves giving of traffic and way finding details to a tourist

Explanation:

Visual communication can improve the message that is transmitted with the speaker through the transmission of additional concepts to the message, such as gestures, body expressions or communicative looks, which add meanings to the words emitted.

Today, visual communication is extremely developed and is designed to perform several tasks at once. Communication through visual images is progressing thanks to the development of new technologies. Visual communication is one of the key components of modern media and social media.

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Answer:

2.62 Ω

Explanation:

For a series motor, the field resistance is in series with the armature resistance. The back emf (e) is given by:

[tex]V=E_b+I_A(R_a+R_f)\\\\Where\ V\ is\ the\ terminal\ voltage, R_a=armature\ resistance,R_f=field \ resistance\\\\Given: V=415\ V,R_a=0.5 \Omega, R_f=0.25 \Omega,I_a=30A\\\\415=E_{b1}+30(0.5+0.25)\\\\415=E_{b1}+22.5\\\\E_{b1}=415-22.5=392.5V[/tex]

For a back emf of 392.5 V, the speed is 1000 rpm.

Speed is directly proportional to back emf. It is given as:

[tex]Nk\phi=E_b\\\\N_1k\phi_1= E_{b1}\\\\N_2k\phi_2= E_{b2}\\\\N_1=1000\ rpm, N_2=800\ rpm, E_{b1}=392.5\\\\\frac{E_{b1}}{E{b2}}= \frac{k\phi_1N_1}{k\phi_2N_2}\\\\\frac{E_{b1}}{E{b2}}= \frac{\phi_1N_1}{\phi_2N_2}\\But\ \phi\ is\ directly\ proportional\ to I_a\\\\\frac{E_{b1}}{E{b2}}= \frac{kI_{a1}N_1}{kI{a2}N_2}\\\\I_{a1}=I_{a2}\\\\\frac{E_{b1}}{E{b2}}= \frac{N_1}{N_2}\\\\E_{b2}=\frac{E_{b1}N_2}{N_1}=\frac{392.5*800}{1000}=314\ V[/tex]

Let the added resistance be R

[tex]V=E_{b2}+I_A(R_a+R_f+R)\\\\415=314+30(0.5+0.25+R)\\\\415=314+22.5+30R\\\\415=336.5+30R\\\\30R=78.5\\\\R=2.62\Omega[/tex]