Which of the following statements about resistance is TRUE?

At an impaired driver checkpoint, the time required to conduct the impairment test varies (according to an exponential distribut

professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

Average time required = 60 seconds

Therefore,

The maximum capacity that can be accommodated on the system, μ = 60 veh/hr

Average Arrival rate, λ = 30 vehicles per hour

Now,

The average time spent by the vehicle is given as

⇒ $\frac{1}{\mu(1-\frac{\lambda}{\mu})}$

thus,

on substituting the respective values, we get

Average time spent by the vehicle = $\frac{1}{60(1-\frac{30}{60})}$

or

Average time spent by the vehicle = $\frac{1}{60(1-0.5)}$

or

Average time spent by the vehicle = $\frac{1}{60(0.5)}$

or

Average time spent by the vehicle = $\frac{1}{30}$ hr/veh

or

Average time spent by the vehicle = $\frac{1}{30}\times60$ min/veh

[ 1 hour = 60 minutes]

thus,

Average time spent by the vehicle = 2 min/veh

Hence,

Option (d) 2 min/veh

7 0

3 years ago

AC motor characteristics require the applied voltage to be proportionally adjusted by an AC drive whenever the frequency is chan

Margarita [4]

The answer is false

6 0

3 years ago

Read 2 more answers

The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm

aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

$T_max =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=14.5MPa

$T_{min} =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=10.3MPa

7 0

2 years ago

A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an

Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here thermodynamic equation that

energy equation that is

$h1 + \frac{v1}{2} + q = h2 + \frac{v2}{2} + w$

put here value with

turbine is insulated so q = 0

so here

$3015.4 *1000 + \frac{10^2}{2} = 2431.7 * 1000 + \frac{90^2}{2} + w$

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0

3 years ago

For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea

miskamm [114]

Answer:

$k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

where $r_1$ and $r_2$ be the inner radius, outer radius of the annalus.

Explanation:

Let $r_1$ , $r_2$ and $L$ be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, $T_1=30^{\circ}C\\$ and at the outer surface, $T_2=40^{\circ}C$ .

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

$40 W/m^2$ .

$\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40$

$\Rightarrow q=2.4\pi L \;W$

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

$q=-kA\frac{dT}{dr}\;\cdots(i)$

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, $A=2\pi rL\;\cdots(ii)$

Variation of temperature w.r.t the radius of the annalus is

$\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}$

$\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)$

Putting the values from the equations (ii) and (iii) in the equation (i), we have

$q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}$

$\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}$

$\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)}$ [as $q=2.4\pi L$ , and $T_2-T_1=10 ^{\circ}C$ ]

$\Rightarrow k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0

3 years ago