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Natalija [7]
3 years ago
13

A large retirement village has a total retail employment of 120. All 1600 of the households in this village consist of two nonwo

rking family members with household income of $20,000. Assuming that shopping and social/recreational trip rates both peak during the same hour (for exposition purposes), predict the total number of peak-hour trips generated by this village using the trip generation models of:_________.
Number of peak hour vehicle based shopping trips per household 0.12
+0.09(household size)
+0.011 (annual house income in thousands of dollors)
-0.15(retail employees in household s neighborhood, in hunreds)
Number of peak hour vehicle based social/recreactional trips per household
= 0.04 + 0.018 (household size)
+0.009(annual house income in thousands of dollors)
+0.16( number of nonworking household members)

Engineering
1 answer:
polet [3.4K]3 years ago
5 0

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

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The emissivity of galvanized steel sheet, a common roofing material, is ε = 0.13 at temperatures around 300 K, while its absorpt
Step2247 [10]

Answer:

759.99W/m²

Explanation:

Question: If the temperature of the sheet is 77C,what is the incident solar radiation on aday with Tinf= Tsurr= 16°C?

Given

Energy Equation of the Gas

αs * Gs * A + h * A * (T inf - Tg) + εσA (Tsurr⁴- Tg⁴) = 0

Where σ= 5.67 *10^-8 W/m²K⁴ (Stefan-Boltzmann constant)

ε = 0.13 (Emisivity)

αs = 0.65 (Absorptivity for solar radiation)

h = 7W/m²K⁴

Tg = 77 + 273.15K = 350.15K

T inf = 16 + 273.15 = 288.15K

T surr= T inf = 288.15

Substitute the above values in the Gas Equation, we have

0.65 * Gs * A + 7 * A * (288.15 - 350.15) + 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴) = 0

0.65 * Gs * A = - 7 * A * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴)

A cancels out, so we are left with

0.65 * Gs = - 7 * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * (288.15⁴ - 350.15⁴)

0.65Gs = 434 - 0.7372 * 10^-8(−8,137,940,481.697)

0.65Gs = 434 + 0.7372 * 81.37940481697

0.65Gs = 493.992897231070284

Gs = 493.992897231070284/0.65

Gs = 759.9890726631850

Gs = 759.99W/m² ------- Approximated

3 0
3 years ago
Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the undergr
maksim [4K]

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

\eta_t = The efficiency of the pump = 78%

\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}

Q_{max} = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump Q_{max} ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

6 0
2 years ago
500 flights land each day at San Jose’s airport. Assume that each flight has a 5% chance of being late, independently of whether
BARSIC [14]

Answer:

a.0.0199

b.0.1765

c.0.0785

d.0.1268

e.Yes

Explanation:

It is given that   X follows a  Binomial distribution with (n= 500, p = 0.05)

The  probabilities  are computed using the EXCEL .

a) The required probability here is:

P(X less of equal to  15)

= binom.dist(15,500,0.05,TRUE)

=0.0199

Therefore the probability is 0.0199 .

b) The required probability here is:

P(X greater or equal to 30) = 1 - P(X less or equal to  29)  

=1 - binom.dist(29,500,0.05,TRUE)

=0.1765

Therefore the probability is 0.1765

c) P(X = 26 )

= binom.dist(26,500,0.05,FALSE)  

=0.0785

Therefore the probability is 0.0785

d) The required probability here is computed as:

P(10 less or equal to X less or equal to 20 ) = P(X less or equal to 19) - P(X less or equal to 10)

= binom.dist(19,500,0.05,TRUE) - binom.dist(10,500,0.05,TRUE)

=0.1268

Therefore the probability 0.1268

e) Yes . Therefore the probability because that is the assumption used to apply binomial distribution .

6 0
3 years ago
Your study space does not need to be quiet as long as you can ignore any noise coming from the space true or false?
Makovka662 [10]

Answer:

False

Explanation:

When you're studying, you need to make sure that you can focus properly. This means that you shouldn't be hungry or too full and that you should be well-rested, in a quiet room with good lighting and no distractions. Noise is never good when you need to memorize something. Some people can partially ignore it as long as it isn't too loud, but it will begin to bother them eventually. That's why it's better to study in a quiet room.

3 0
2 years ago
Read 2 more answers
A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of
OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

7 0
2 years ago
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