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Bess [88]
3 years ago
14

A driver counts 21 other vehicles using 3 EB lanes on one section of I-80 between her rented car and an overpass ahead. It turne

d out to be 0.78 mile to the overpass from the point at which she started her vehicle count. The vehicles were moving at approximately 52 mph at the time. Traffic was distributed evenly over the three EB lanes.
Required:
a. Calculate the vehicle density per lane (excluding the driver's rental vehicle) on that stretch of EB I-80.
b. What was the flow rate for that section of I-80?
Engineering
1 answer:
SVEN [57.7K]3 years ago
6 0

Answer:

vehicle density =  28.205 veh/mile

flow rate =  0.909 veh/hr

Explanation:

given data

count n = 21

distance = 0.78 miles

speed = 52 mph

solution

we get here vehicle density that is express as

vehicle density = n ÷ distance    ...............1

vehicle density = ( 21 + 1 )  ÷ 0.78

vehicle density  k =  28.205 veh/mile

and

now we get here flow rate that is express as

flow rate = k × vs    .................2

flow rate = 28.205 × ( 52 × 0.00062 ÷ 1m )

flow rate =  0.909 veh/hr

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goblinko [34]

Answer:

45.3 MN

Explanation:

The forging force at the end of the stroke is given by

F = Y.π.r².[1 + (2μr/3h)]

The final height, h is given as h = 100/2

h = 50 mm

Next, we find the final radius by applying the volume constancy law

volumes before deformation = volumes after deformation

π * 75² * 2 * 100 = π * r² * 2 * 50

75² * 2 = r²

r² = 11250

r = √11250

r = 106 mm

E = In(100/50)

E = 0.69

From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula

F = Y.π.r².[1 + (2μr/3h)]

F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]

F = 35.3 * [1 + 0.2826]

F = 35.3 * 1.2826

F = 45.3 MN

7 0
4 years ago
In python, how would I randomize numbers and insert them into a file?
BabaBlast [244]
Random integer values can be generated with the randint() function. This function takes two arguments: the start and the end of the range for the generated integer values. Random integers are generated within and including the start and end of range values, specifically in the interval [start, end].
7 0
3 years ago
Find the resistance of a circuit that draws 4 amperes with 8 volts applied?
vodka [1.7K]

Answer:

2 ohms

Explanation:

V = I * R

8 = 4 * R

8 / 4 = R

R = 2 ohms

5 0
3 years ago
Recall that I format instruction include Rs and Rt fields. When executing an LW instruction, is the memory address computed usin
kipiarov [429]

Answer:

No, only Rs

Explanation:

The instructions lw and sw use the instruction's Rs and immediate fields to compute the data memory address that will be accessed by

3 0
4 years ago
Design a rectangular metallic waveguide to be used for transmission of electromagnetic power at 2.45 GHz. This frequency should
Vera_Pavlovna [14]

Answer:

A) 1.4 *10^11 watts

B)  41.42 ≈ 41 TIMES

Explanation:

Designing a rectangular metallic wave guide using the given data

Electromagnetic power = 2.46 GHz also at the middle of operating frequency

A) Design an air-filled guide to meet the given specifications.

operating frequency range =  C / αa < f < C / a

2.45 GHz = \frac{\frac{C}{ba}+ \frac{c}{a}  }{2}  

The given frequency middle at the middle of operating frequency range

= 4.9 GHz = \frac{c + 2c }{ba}  = 3C / βa

α = \frac{3*3*10^{10} }{2*4.9*10^9}  = 45/4.9 = 9.18 cm

note: to operate in dominant mode aspect ratio should be  b = α/2

therefore b = 4.59 cm

Also Maximum power can be carried by wave guide only in dominant mode

i.e TE10 mode

power carried = I E I^2ab / 4Zte   using this formula

ZTE = impedance when operated in TE mode = \sqrt[n]{1-(\frac{Fc}{f} )^{2} }

Fc = cutoff frequency = (3*10^16) / (2*9.18) = 1.6GHz

F = operating frequency = 2.45 GHz

n = freespace impedance = 377 ohms

input all the given values back to ZTE  equation

ZTE = 285 ohms

power carried = \frac{|2*10^6|^{2}* 9.18 * 4.59 }{4 * 285}  =  4*10^12 * 0.036

THEREFORE power carried 1 = 1.4 *10^11 watts

B) The dielectric materials given data/parameters

∈ = 2.5 ∈o   ∪ = ∪o

breakdown field = 10^7

free space impedance  n = \sqrt{\frac{u}{e} } = \sqrt{\frac{UoUr}{EoEr} }

therefore for the given dielectric n = \sqrt{\frac{Uo}{Eo} } \sqrt{\frac{1}{2.5} } = \frac{377}{\sqrt{2.5} }     n = 238.43

ZTE = \sqrt[n]{1-(\frac{1.6}{2.45} )^{2} }    

therefore ZTE = 180.56 ohms

power carried 2 = \frac{|10^7|^2*9.18*4.59}{4*180.56}  = 58*10^{11}  N

To calculate the number of time power can be transmitted by the waveguide = power carried 2 / power carried 1

=  58*10^11 / 1.4*10^11  = 41.42 ≈ 41

3 0
3 years ago
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