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IrinaVladis [17]
3 years ago
7

the AADT for a section of suburban freeway is 150000 veh/day. Assuming this is an urban radial facility, what range of direction

al design hour volumes would be expected?
Engineering
1 answer:
igomit [66]3 years ago
7 0

Answer:

design hour volumes will be 4000 to 6000

Explanation:

given data

AADT  = 150000 veh/day

solution

we get here design hour volumes that is express as

design hour volumes  = AADT × k × D    ..............1

here k is factor and its  range is 8 to 12 % for urban

and D is directional distribution i.e traffic equal divided by the direction

so here design hour volumes will be 4000 to 6000

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What happens to a commercial airline at cruising altitude if the pilot does not touch the throttles?
jolli1 [7]

Answer:

it slows down

Explanation:

8 0
3 years ago
John has just graduated from State University. He owes $35,000 in college loans, but he does not have a job yet. The college loa
IRISSAK [1]

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= 35000\times (\frac{F}{P}, 4 \ percent,5 )

= 35000\times 1.216 7

= 42584.50

Now,

time (t) will be:

= 5\times 12

= 60 \ monthly \ payments

So, monthly payment will be:

= 42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )

= 42584.50\times 0.0193

= 821.88

6 0
3 years ago
A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic
Roman55 [17]

Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

5 0
3 years ago
Write the following decorators and apply them to a single function (applying multiple decorators to a single function): 1. The f
natita [175]

Answer:

Complete question is:

write the following decorators and apply them to a single function (applying multiple decorators to a single function):

1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <strong> and </strong> to the argument of the decorator. The return value of the wrapper should look like: return “<strong>” + func() + “</strong>”

2. The decorator will return the wrapper per usual.

3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <em> and </em> to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “<em>” + func() + “</em>.

4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.

5. Apply both decorators (by @ operator to greetings()).

6. Invoke the greetings() function and capture the result.

Code :

def strong_decorator(func):

def func_wrapper(name):

return "<strong>{0}</strong>".format(func(name))

return func_wrapper

def em_decorator(func):

def func_wrapper(name):

return "<em>{0}</em>".format(func(name))

return func_wrapper

@strong_decorator

@em_decorator

def Greetings(name):

return "{0}".format(name)

print(Greetings("Hello"))

Explanation:

5 0
3 years ago
A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by
gregori [183]

Answer:

(a) T = W/2(1-tanθ)  (b) 39.81°

Explanation:

(a) The equation for tension (T) can be derived by considering the summation of moment in the clockwise direction. Thus:

Summation of moment in clockwise direction is equivalent to zero. Therefore,

T*l*(sinθ) + W*(l/2)*cosθ - T*l*cosθ = 0

T*l*(cosθ - sinθ) = W*(l/2)*cosθ

T = W*cosθ/2(cosθ - sinθ)

Dividing both the numerator and denominator by cosθ, we have:

T = [W*cosθ/cosθ]/2[(cosθ - sinθ)/cosθ] = W/2(1-tanθ)

(b) If T = 3W, then:

3W = W/2(1-tanθ),

Further simplification and rearrangement lead to:

1 - tanθ = 1/6

tanθ = 1 - (1/6) = 5/6

θ = tan^(-1) 5/6 = 39.81°

8 0
3 years ago
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