Answer: the distance in advance of the curve the sign must be located to ensure that vehicles have sufficient distance to decelerate safely is 292.07 ft
Explanation:
first we calculate the distance to place sign board in advance to curve for decelerating the speed
d = 1.47Si t + [ (Si² - Sf²) / 30(0.348 ± 0.01G)
where d is the safe stopping distance, initial speed is Si (60 mi/hr ), time reaction is t (2.5s), Sf is the final speed (40 mi/hr), G is the grade (0).
so we substitute
d = (1.47 × 60 × 2.5) + [ (60² - 40²) / 30(0.348 ± 0)
= 220.5 + ( 2000/10.44)
= 220.5 + 191.57
= 412.07 ft
Now giving that a warning sign is clearly visible at a distance of 120 ft
optimum safe distance will be
d = minimum distance - sign visible distance
d = 412.07 - 120
d = 292.07 ft
therefore the distance in advance of the curve the sign must be located to ensure that vehicles have sufficient distance to decelerate safely is 292.07 ft
An ideal Otto Cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression process, air is at 98kPa and 20C. The pressure is doubled during the constant volume heat addition process. Assuming constant specific heats, determine:
Answer: hello your question is incomplete below is the complete question
An ideal Otto Cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression process, air is at 98kPa and 20C. The pressure is doubled during the constant volume heat addition process. Assuming constant specific heats, determine:the amount of heat transferred to the air
answer : 609.804 kj/kg
Explanation:
Given data:
compression ratio (r)= 9.2
pressure given(p1) = 98 kPa
Initial temperature = 20 + 273 = 293 k
pressure during constant volume heat addition process = 2p1
note : specific heat at constant pressure and specific heat at constant volume varies with temperature
we use T = 300k because it is closest to T1 = 293 k
hence at T = 300 K ( ideal gas properties of air )
[tex]u_{1}[/tex] = 214.07 Kj/kg
[tex]v _{r1}[/tex] = 621.2
To get [tex]v_{r2}[/tex] = [tex]v_{r1} * \frac{v_{2} }{r}[/tex] = 621.2 * 1 / 9.2 = 67.52
ALSO at [tex]v_{r2}[/tex] = 67.52 ( from ideal gas properties )
[tex]u_{2}[/tex] = 518.9 kj/kg
T2 = 708.32 k
next we apply the gas equation
[tex]\frac{p1v1}{T1} = \frac{p2v2}{T2}[/tex]
hence p2 = (9.2) * [tex]\frac{708.32}{293} * 98[/tex] = 2179.59 kpa
to determine T3 due to the constant volume heat addition
[tex]\frac{T3}{T2} = \frac{P3}{P2}[/tex]
Hence T3 = p3/p2 * T2 = 2( 708.32 ) = 1416.64 k
At T3 = 1416.64 k ( from ideal gas properties )
[tex]u_{3}[/tex] = 1128.704 kj/kg
[tex]v_{r3}[/tex] = 8.592
Determine the amount of heat transferred to the air
[tex]q_{in} = ( u_{3} - u_{2} )[/tex]
= ( 1128.704 - 518.9 )
= 609.804 kj/kg
A 75 ! coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 j75 !. If the relative permittivity of the line is 2.56 and the frequency is 3.0 GHz, find the input impedance to the line, the reflection coefficient at the load, the reflection coefficient at the input, and the SWR on the line
Answer:
4.26
Explanation:
The wavelength λ is given by:
[tex]\lambda=v/f=c/nf\\c=speed\ of\ light=3*10^8m/s,f=frequency=3*10^9Hz,n=permittivity=2.56\\\\\lambda=3*10^8/(2.56*3*10^9)=0.0625\ m\\[/tex]
Phase constant (β) = 2π/λ
βl = 2π/λ × l
l = 2 cm = 0.02 m
βl = 2π/0.0625 × 0.02=2.01 rad = 115.3°
1 rad = 180/π degrees
[tex]Z_L=load\ impedance=37.5+j75\\\\Z_o=characteristic impedance = 75\ ohm\\\\\tilde {Z_L}=Z_L/Z_o=37.5+j75/75=0.5+j[/tex]
[tex]\tilde {Z_{in}}=\frac{\tilde {Z_{L}}+jtan\beta l}{1+j\tilde {Z_{L}}tan\beta l}=\frac{0.5+j+jtan(115.2)}{1+j(0.5+j)tan(115.2)}=0.253-j0.274\\ \\Z_{in}=Z_o\tilde {Z_{in}}=75(0.253-j0.274)=19-j20.5\\\\\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_o}=\frac{37.5+j75-75}{37.5+j75+75}=0.62\angle 83^o\\\\\Gamma_{in}=\frac{Z_{in}-Z_0}{Z_{in}+Z_o}=\frac{19-j20.5-75}{19-j20.5+75}=0.62\angle -147^o\\\\VSWR=\frac{1+\rho}{1-\rho} =\frac{1+0.62}{1-0.62}=4.26[/tex]
Determine the angles made by the vector V = - 36i + 15j with the positive x- and y-axes. Write the unit vector n in the direction of V.
Answer:
157.38° CCW from +x axis
67.38° CCW from +y axis
n = (12/13)i +(5/13)j
Explanation:
The reference angle is ...
arctan(15/36) ≈ 22.62° . . . . CW from the -x axis
The signs of the component vectors indicate this is a 2nd-quadrant vector, so the angles of interest are ...
157.38° CCW from +x axis
67.38° CCW from +y axis
The unit vector will be ...
n = cos(157.38°)i +sin(157.38°)j
n = (12/13)i +(5/13)j
Given the following MATLAB statement: ( 3 + 2 ) / 5 * 4 + 5 ^ 2 In what order will these operations be done?
Answer:
first is the parentheses, (3+2)=5 next is the exponent 5^2=25, next is the division 5 / 5 = 1, then the multiplication 4*1=4 and then you add 4+25=29. so the answer is 29.
what add the most carbon dioxide to the atmosphere
Answer:
Volcanic outgassing
The burning of fossil fuels (coal, oil, gasoline, natural gas), for heating, power generation and transport
Some industrial processes such as cement making.
Explanation:
What would be the required voltage of an energy source in a circuit with a current of 10.0 A and a resistance of 11.0 Ω?
Answer:
110 V
Explanation:
V = IR
V = (10.0 A)(11.0 Ω) = 110 volts
Three loads are connected in parallel across a single-phase source voltage of 480 V (RMS). Load 1 absorbs 10 kW and 8 kVAR; Load 2 absorbs 5 kVA at 0.9 power factor leading; Load 3 absorbs 12 kW at 0.95 power factor leading. Calculate the equivalent impedance, Z, for the three parallel loads, for two cases: (a) Series combination of R and X, and (b) parallel combination of R and X. Problem
Answer:
a) ZT = [ 38.786 ∠-4.28° ] ohms
b) ZT = [ 8.673 ∠4.06° ] ohms
Explanation:
Given that;
Voltage V = 480V
three loads of 1) 10Kw, 8 kVAR
2) 5 kVA, 0.9 power factor lagging
3) 12kW, 0.95 pf leading
Now for load1
Active power P = V^2 / R
Resistance R1 = V^2 / P
= (480 * 480) / 10*1000 = 23.04 OHMS
Reactive power Q = 8kVAR
let x be reactance
Q = V^2 / x
8 * 1000 = (480 * 480) / x
x1 = j28.8 ohms (inductive)
Load 2
Active power = kVA * nf = 5 * 9 = 4.5 Kw
Reactive power = sqrt(S^2 - P^2)
= Sqrt (5^2 - 4.5^2)
= 2.179 kVAR
now Resistance R2 = V^2 / P2 = (480 * 480) / 4.5*10^3
= 51.2 ohms
Reactance X2 = V^2 / Q2 = (480 * 480) / 2.178*10^3
X2 = 105.75 ohms
x2 = -j105.75 ohms (capacitive)
Load 3
Active power = 12kW AND PF = 0.95
so
kVA rating = 12/0.95 = 12.63 kVA
reactive power Q3 = sqrt (12.63^2 - 12^2 )
= 3.94 kVAR
Resistance R3 = V^2/P3 = (480*480) / 12*10^3
= 19.2 ohms
Reactance x3 = V2 / Q3 = (480 * 480) / 3.94*10^3
x3 = 58.47 ohms
x3 = -j58.47 ohms (capacitive)
NOW
a)
series combination of R and X
1/ZT = [1/(R1 + jx1)] + [1/(R2-jx2)] + [1/(R3-jx3)]
we substitute
1/ZT = [1/(23.04 + j28.8)] + [1/(51.2 - j105.74)] + [1/(19.2-58.48)]
1/ZT = 0.02571 + j1.92688*10^-3
ZT = 38.67 -j2.897 (total impedance)
so ZT = [ 38.786 ∠-4.28° ] ohms
b)
parallel combination of R AND X
1/ZT = 1/R1 + 1/jx1 + 1/R2 + 1/(-jx2) + 1/R3 + 1/(-jX3)
1/ZT = 1/23.04 + 1/j28.8 + 1/51.2 + 1/(-j105.74) + 1/19.2 + 1/(-58.48)
1/ZT = 0.1150 - j0.008164
ZT = 8.65 +j0.6141
so ZT = [ 8.673 ∠4.06° ] ohms
If you are driving a 30-foot
vehicle at 55 mph, you should
leave how many seconds of
following distance?
Answer:
3 sec
If you are driving a 30-foot vehicle at 55 mph, how many seconds of following distance should you allow? 30ft truck. = 3 sec. Since the truck is over 40 mph.
Explanation:
Mechanical clips are used to close bags and keep things together. Investigate the design of various paper clips, hair clips, and potato chip bag clips. Choose three (3) different types of clips. Discuss what you think are important design parameters. Discuss the advantages and disadvantages associated with each of the three (3) designs.
Answer:
Hair clip: this used to hold human hairs together especially long hairs
The Design parameters to be considered are : number of teeth, durability of the material to be used and also the width ( how wide the clip can open )
Advantages : very easy to use, holds hairs tight
Disadvantages : Expensive and material used is sometimes brittle
paper clips : This used to hold papers together it is made of a thin elastic wire
The design parameters to be considered are : length of the wire, material to be used and the elasticity of the material
advantages : it is quite cheap and good strength
disadvantages :it can hold a limited amount of papers together
potato chip bag clips: These clips are used on potato chip bags to ensure that the bags are air tight
The design parameters to be considered are : Elasticity of the material to be used , and size
Advantages : it helps to prevent the potato chips from getting exposed to air
disadvantages : Not very popular and its quite expensive
Explanation:
Hair clip: this used to hold human hairs together especially long hairs
The Design parameters to be considered are : number of teeth, durability of the material to be used and also the width ( how wide the clip can open )
Advantages : very easy to use, holds hairs tight
Disadvantages : Expensive and material used is sometimes brittle
paper clips : This used to hold papers together it is made of a thin elastic wire
The design parameters to be considered are : length of the wire, material to be used and the elasticity of the material
advantages : it is quite cheap and good strength
disadvantages :it can hold a limited amount of papers together
potato chip bag clips: These clips are used on potato chip bags to ensure that the bags are air tight
The design parameters to be considered are : Elasticity of the material to be used , and size
Advantages : it helps to prevent the potato chips from getting exposed to air
disadvantages : Not very popular and its quite expensive
True or False? A constricting nozzle is used
to create, concentrate, and direct the high-
velocity plasma.
Answer:
True
Explanation:
Principles of plasma arc cutting, Uses a constricting nozzle to create, concentrate, and direct the high-velocity plasma. Plasma gas is always used
in plasma arc cutting When shielding gas is also used, the process is called dual flow plasma arc cutting. I hope this helps.
The Release Train Engineer is a servant leader who displays which two actions or behaviors?
Explanation:
The Release Train Engineer (RTE) has the main work of supporting as well as coaching the Agile Release Train (ART). They are capable of steering ART successfully and to navigate the complexity in delivering the software in large and inter-functional environments.
They serve the scrum master and coach teams to improve on the results.
The two actions or behaviors of the RTE are :
1. They try to create an environment of the mutual influence.
2. Listens and also supports the teams in problem identification as well as decision-making.
The behaviors or actions that the release train engineer does include:
Operating within the lean budget.Facilitating demosThe release train engineer refers to a servant leader who is responsible for facilitating program-level processes and executes them.
The release train engineer is also responsible for driving continuous development, managing risks, and escalating impediments. Some of their actions include operating within the lean budget and facilitating demos.
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