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
Ten dollars per hour is about how much income per year
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.
A helical compression spring is to be made with oil-tempered wire ol 4 mm diameter with a spring index of C = 10 The spring is to operate inside a hole, so buckling is not o problem and the ends can be left plan The free length of the spring should be 80 mm. A force of 50 N should deflect the spring 15 mm (a) Determine the spring rate (b) Determine the minimum hole diameter tot the spring to operate in (c) Determine the total number of coils needed (d) Determine the solid length.
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:
d=4mmC=10mm; D/d = mean coil diameter / wire diameter = 10; D = 10 * 4 = 40mm.Free length of the spring = 80mmForce (f)= 50Nδ=15mma) 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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Drag each tile to the correct box.
Arrange the types of movie discs in terms of the amount of content they can hold.
DVD
Blu-ray
CD
->
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:
In a four-stroke engine, the piston rises in the cylinder, which triggers the _______ stroke.
intake
compression
exhaust
power
Answer: It is power stroke
How can visual communication enhance the message conveyed with the speaker? Cite with a particular situation where in it is best to use visual communication
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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A concrete column has a diameter of 390 mm and a length of 2.8 m . If the density (mass/volume) of concrete is 2.45 Mg/m3, determine the weight of the column in pounds.
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,
Diameter of column is 390mm or 0.39m.Length (l) of column is 2.8m.Density of column is 2.45 Mg/m3According 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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Since most architects have at least a bachelor's degree, they do not require on-the-job training.
O True
O False
Answer:
The answer is false, I hope this helps. :3
Answer:
False
Explanation:
I took the quiz and it was correct
The repair and rehabilitation of damaged or spalled concrete is done by removing the:____________A. Loose concreteB. CollapsingC. BreakageD. Bonding
Answer:
A. Loose concrete
Explanation:
Concrete is a solid composite structural material.It is the most used component in setting up a structure. Concrete is known for its very good ability to bear load, but can sometimes fail, or can be damaged. To rehabilitate or repair a concrete, the loose concrete must first be removed, so that the surface can be cleaned and well set before the material used for repair is applied on it.
If a transistor is operating with 5 mA of Collector Current, 100uA of Base Current and VCE = 10 V. What is the power dissipation?
Answer:
well its 31.4226804124
Explanation:
the amount of phase shift between input and output signal is important when measuring ____ circuit
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.
In a voltage-divider biased npn transistor, if the upper voltage-divider resistor (the one connected to VCC) opens:______.
a) the transistor goes into cutoff.
b) the transistor goes into saturation.
c) the transistor burns out.
d) the supply voltage is too high.
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 )
What would happen if an exposed film was accidentally placed in the fixer before being placed in the developer
The autorotation spin characteristics of a straight-wing aircraft are induced by Group of answer choices
Answer:
More Drag on the down going wing and More Lift on the up going wing
Explanation:
The autorotation spins of blades used in airborne wind energy technology sectors help drive and move the winds and water propeller-type turbines or shafts of generators to produce electricity at altitude and transmit the electricity to earth through conductive tethers.
Sometimes autorotation takes place in rotating parachutes, kite tails. Etc.
As a result, more Drag usually induces the autorotation spin characteristics of a straight-wing aircraft on the downgoing wing and More Lift on the up-going wing.
1. Represent each of the following combinations of units in the correct SI form using the appropriate prefix: (a) μMN, (b) N/μm, (c) MN/ks2, and (d) kN/ms.
2. Represent each of the following quantities in the correct SI form using the appropriate prefix: (a) 0.000431 kg, (b) 35.3(103) N, (c) 0.00532 km.
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).
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)
Which of the following answer options can damage flexible and extension cords?
Staples
Door or window
Fastenings
Aging
Abrasion from adjacent
(Select all that apply)
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.
What is Extension cords?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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The subsequent result of the
system to the input is known
as
Select one:
A.Response
B.Command
C.Process Control
D.Process Controller
as
10. The repair order is a legal document because
A. the shop owner signs it
B. it's signed by the customer.
C. the manufacturer authorizes it
D. it's paid for by the customer
Who is father of Engineer?
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
5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025
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
Pick up the correct statement from the following:
A. The theory of formation of concrete is based on the phenomena of formation of voids
B. The bulking of sand is taken into account while volumetric proportioning of the aggregates
C. The dry sand and the sand completely flooded with water, have practically the same volume
D. All the above
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.
Copper (Cu) has FCC crystal structure. Atomic mass of Cu is 63.55 gmol-1 and radius of a copper atom is 0.128 nm, mean speed of conduction electrons in Cu is 1.5 × 10^6 m/s and frequency of vibration of the Cu atoms at 27 ° is 4 × 10^12 s-1. Nordheim’s Coefficient of Au in Cu is 5500 nΩ −m. a) Calculate the atomic density of copper b) Estimate the drift mobility of electrons in Cu c) Find the resistivity of pure Cu d) Find the resistivity of Cu if 1 at.% of Au is mixed in Cu. pls solve it
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:
The magnitude of the line voltage at the terminals of a balanced Y-connected load is 660 V. The load impedance is 240-j70 Ω/Φ. The load is fed from a line that has an impedance of 0.5 + j4 Ω/Φ.
A) What is the magnitude of the line current?
B) What is the magnitude of the line voltage at the source?
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
Once the ideas were narrowed down and divided into categories, the group was split into four smaller teams. What phase(s) of the design process was each of these groups responsible for?
A. Shopping
B. Safety
C. Checkout
D. Finding stuff you need
Answer:
Finding stuff you need ( D )
Explanation:
Once an idea is narrowed down and divided into categories, The next line of action would be to create smaller groups that will be brainstorm solutions towards resolving the problem/ implement the idea faced by the team.
The phase of the design process that the smaller teams created will be focused on is to model and build a prototype ( finding stuff you need ) that would help implement the ideas
Under the right conditions, it is possible, due to surface tension,to have metal objects float on water. Consider placing a shortlength of a small diameter steel ( γ = 490 lb/ft3)rod on a surface of water. What is the maximum diameter that therod can have before it will sink? Assume that the surface tensionforces act vertically upward. Note: A standard paper cliphas a diameter of 0.036 in. Partially unfold a paper clip and seeif you can get it to float on water. Do the results of thisexperiment support your analysis?
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
Which of the following statement(s) are true?
A. NAND gates output a 1 when each of its inputs is a 1.
B. All circuits can be created using only NOR gates.
C. A circuit with 4 inputs has 16 possible combinations (truth table lines).
D. You should never tie 2 outputs of two distinct circuits together.
Sam constructs a circuit, connects a lead acid battery of 2 V to a lamp of resistance 3 Ω and places an ammeter across it. What must be the reading of the ammeter?
A.
0.66 A
B.
0.5 A
C.
0.54 A
D.
0.61 A
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.
For the reactions of ketone body metabolism, _______.
A. NADH is produced by catabolism of D‑β‑hydroxybutyrate.
B. Liver lacks thiolase and therefore cannot use ketone bodies as a fuel.
C. Conversion of 2 acetyl-CoA to acetoacetyl-CoA is accompanied by hydrolysis of ATP to AMP and PPi.
D. The enzymes that catalyze biosynthesis of ketone bodies are found in the cytosol of hepatocytes.
Answer:
Is
Explanation:
The Ksp for Ag2CrO4 is 9.0 x 10-12. If 200 mL of 0.0050 M AgNO3 is combined with 300 mL of 0.0020 M K2CrO4, will a precipitate form?
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
A series motor runs at 1000 r.P.M. When the voltage is 415 V and the current is 30 A. The armature resistance is 0.5Ω and the series field resistance is 0.25 Ω. Determine the resistance to be connected in series to reduce the speed to 800 r.P.M with the same current.
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]
Trevor typically works on the highways doing maintenance. Who is most likely his employer? a large private company himself the government a small private company
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 ;)