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3 years ago

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Assume you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 30 mi/h, all

vehicles in lane 2 are traveling at 45 mi/h, and all vehicles in lane 3 are traveling at 60 mi/h. There is also a constant spacing of 0.5 mile between vehicles.
If you collect spot speed data for all vehicles as they cross your observation point, for 30 minutes,
what will be the time- mean speed and space-mean speed for this traffic stream?

1 answer:

garik1379 [7]3 years ago

5 0

Answer:

Explanation:

Given data;

In line 1, v1 = 30mi/hr

in line 2, v2 = 45mi/hr

in line 3, v3 = 60mi/hr

therefore time mean speed = v1 + v2 + v3 /n

= VT = 45mi/hr

space mean speed ; Vs

harmonic mean = 1/V = 1/v1 + 1/v2 + 1/v3

V = 13.85mi/hr

Hence Vs = V x n = 3 x 13.85 = 41.55mi/hr

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Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th

Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine = $1 - \frac{Tl}{Th} \\$

where Th =temperature at which the heat enters the engine

Tl is the temperature of the environment

since both engines have the same thermal capacities <em> $n_{th}$ </em> therefore $n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\$

We have now that

$\frac{-1400}{T}+\frac{T}{300}=0\\$

multiplying through by T

$-1400 + \frac{T^{2} }{300}=0\\$

multiplying through by 300

- $420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000} \\T=648.07k$

The temperature T= 648.07k

5 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.

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3 years ago

How do you build a engine best anwser will get marked as brainlist

marysya [2.9K]

Answer:

It is best to soak the new lifters in engine oil overnight in order to build a car engine from scratch, regardless of its length. Ensure all rings of the pistons are installed correctly. You should flip the block over and carefully install the top half of the main rod and bearing parts. Get the block out of your way and flip it over again. By flipping over the block, you’ll see the full effect. Ensure that the lifters are installed correctly.

Explanation:

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2 years ago

On highways, the far left lane is usually the _____. A. emergency lane B. merge lane C. slowest D. fastest

slamgirl [31]

On highways, the far left lane is usually the<u> fastest</u> moving traffic.

Answer: Option D.

<u>Explanation:</u>

For the most part, the right lane of a freeway is for entering and leaving the traffic stream. It is an arranging path, for use toward the start and end of your interstate run. The center paths are for through traffic, and the left path is for passing. On the off chance that you are not passing somebody, try not to be driving in the left path.

Regular practice and most law on United States expressways is that the left path is saved for passing and quicker moving traffic, and that traffic utilizing the left path must respect traffic wishing to surpass.

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4 years ago

Read 2 more answers

6.3.9 A coin was tossed n = 1000 times, and the proportion of heads observed was

4vir4ik [10]

Answer:

No

Explanation:

51 / 100 = 510 / 1000

Chance of getting a head is 1 / 2 of total throws

= 1 / 2 × 1000

= 500 is the probability

and the number of heads was just 10 more the the probability...if the was a greater gap, there would be evidence to say the coin is unfair

4 0

4 years ago