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Natalija [7]
4 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]4 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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Pouring molten aluminum into a mold and allowing it to cool forms?
neonofarm [45]

Answer:it forms a molten mold that makes it hard to be able to smash something into it then make something like a key

Explanation:

7 0
3 years ago
Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)
ArbitrLikvidat [17]

Answer:

\mathbf{P_x =25 \ watts}

\mathbf{x_{rmx} = 5 \ unit}

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) +  sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal  Asin (ωt), Power can be expressed as:

P= \dfrac{A^2}{2}

For the number of sinosoidial signals;

Power can be expressed as:

P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...

As such,

For x(t), Power  P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}

P_x = \dfrac{25}{2}+ \dfrac{25}{2}

P_x = \dfrac{50}{2}

\mathbf{P_x =25 \ watts}

For the number of sinosoidial signals;

RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...

For x(t), the RMS value is as follows:

x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }

x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }

x_{rmx }=\sqrt{(\dfrac{50}{2} )}

x_{rmx} =\sqrt{25}

\mathbf{x_{rmx} = 5 \ unit}

8 0
3 years ago
An excited electron in an Na atom emits radiation at a wavelength 589 nm and returns to the ground state. If the mean time for t
11Alexandr11 [23.1K]

Answer:   Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm

                length of the photon emitted: 6.0 m

Explanation:

The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.

Recall the Heisenberg's uncertainty principle:-

                                 ∆t∆E ≈ h ( Planck's Constant)

The transition time ∆t corresponds to the energy that is ∆E

E=h/t = \frac{(1/2\pi)*6.626*10x^{-34} J.s}{20*10x^{-9} } = 5.273*10x^{-27} J =  3.29* 10^{-8} eV.

The corresponding uncertainty in the emitted frequency ∆v is:

∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)=  7.958 × 10^6 s^-1

To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate

                                                    λ = c/v

                                                    dλ/dv = -c/v² = -λ²/c

Therefore,

      ∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)

                                 =  9.20 × 10⁻¹⁵ m or 9.20 fm

     The length of the photon (<em>l)</em> is

l = (light velocity) × (emission duration)

  = (3.0 × 10⁸  m/s)(20 × 10⁻⁹ s) = 6.0 m          

                                                   

6 0
3 years ago
A 15-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when o
just olya [345]

Answer:

The length of bar will be 2.82 m

Explanation:

Given that

d= 15 mm

r= 7.5 mm

Shear stress = 110 MPa

θ =  30°                                  (30°   = 30°  x π/180°  =0.523 rad)

θ = 0.523 rad

G for steel

G= 79.3 GPa

We know that

\dfrac{\tau}{r}=\dfrac{G\theta }{L}

\dfrac{110}{7.5\times 10^{-3}}=\dfrac{79.3\times 10^3\times 0.523 }{L}

L= 2. 82 m

The length of bar will be 2.82 m

8 0
3 years ago
A universal shift register can shift in both the left-to-right and right-to-left directions, and it has parallel-load capability
soldi70 [24.7K]

Answer:

Explanation:

A unidirectional shift register allows for the capability of shifting in one direction as the name unidirectional implies.

A bidirectional shift register has the capabilities of shifting in both the left to right and right to left directions.

For a Universal shift register, there are possibilities of bidirectional shifts and parallel-load capabilities that have the following properties.

There is the existence of a clear control input whose main  FUNCTION is to set all the register to 0.'

A shift control to both the right and left direction to enable both the shift right operation and shift left operation.

Finally, a parallel-load control whose function is to activate a parallel transfer from input to output.

The diagram for the circuit for such a shift register can be seen in the diagram attached below.

7 0
3 years ago
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