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sergeinik [125]
1 year ago
6

A __________ is a single lane used by drivers to enter and exit a freeway. a) Climbing lane b) Weave lane c) Thru lane d) Offset

lane
Engineering
1 answer:
Rufina [12.5K]1 year ago
6 0

A Weave lane is a single lane used by drivers to enter and exit a freeway.

<h3>What is this lane about?</h3>

Weave lanes are known to be lanes that acts as an entrance and exits for a lot of cars in highways.

Hence, one can say that A Weave lane is a single lane used by drivers to enter and exit a freeway.

Learn more about Weave lane from

brainly.com/question/10828527

#SPJ1

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A BOD test is to be run on a sample of wastewater that has a five-day BOD of 230 mg/L. If the initial DO of a mix of distilled w
Alexxandr [17]

Answer:

Distribution factor P = =38.33

V = 7.826 ml

Explanation:

given details:

BOD =230 mg/l

DO inital = 8.0mg/l

DO final = 2.0mg/l

we know

BOD = [DO inital -DO final] * distribution factor

230 = [8 - 2] D.F

Distribution factor P = \frac{230}{6}

Distribution factor P = =38.33

THE RANGE OF WASTE WATER VOLUME IN 300 ml bottle is

distribution factor = \frac{300}{V}

V = \frac{300}{38.33}

V = 7.826 ml

6 0
2 years ago
. In order to prevent injury from inflating air bags, it is recommended that vehicle occupants sit at least __________ inches aw
scZoUnD [109]
10 inches is the answer
8 0
2 years ago
Read 2 more answers
A motorist enters a freeway at 25 mi/h and accelerates uniformly to 65 mi/h. From the odometer in the car, the motorist knows th
Helga [31]

Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at  

Replacing by the values already known for vo, vf and a, and solving for t, we get:

t = vf-vo /a = (29 m/s – 11.2 m/s) / 2.2 m/s = 8 sec

5 0
2 years ago
A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b
Dennis_Churaev [7]

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions  = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition  Cβ = 87  

we use here lever rule that is

Wα = Wβ   ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

\frac{C_\beta - Co}{C_\beta - Ca}   = 0.5

\frac{87 - 57}{87 - Ca} = 0.5  

solve it we get

Ca = 27

so composition of alpha phase is 27% B

8 0
3 years ago
Which of the following placards allows a driver to park in a handicapped parking space.
malfutka [58]

All of the above ......
4 0
3 years ago
Read 2 more answers
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