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Nitella [24]
3 years ago
9

Drivers with an average of 20/40 vision travel at 55 mph in the curb lane of a freeway, where exit ramps are designed for 25 mph

. What should be the minimum distance of signs with 6-in. letters placed ahead of the exit?
The following information may be useful:
Perception-reaction time 2.5 seconds;
Deceleration rate - 5 ft/sec2:
Road level-1% (downgrade);
Drivers with 20/20 vision can read signs at 60 ft per inch of letter height. .
Engineering
1 answer:
Tom [10]3 years ago
8 0

Answer:

The sign board must be placed 573 ft ahead of the exit.

Explanation:

Distance needed for reducing the speed from 55 mph to 25 mph is given as

d=\frac{v_f^2-v_i^2}{30 \times (\frac{a}{g}-G)}

Here

  • v_f is the velocity at the end which is 25 mph
  • v_i is the velocity at the start which is 55 mph
  • a is the rate of deceleration which is -5 ft/s^2
  • G is the Road grade which is 1% or 0.01
  • g is the gravitational acceleration whose value is 32.2 ft/s^2

d=\frac{v_f^2-v_i^2}{2 \times (\frac{a}{g}-G)}\\d=\frac{25^2-55^2}{30 \times (\frac{-5}{32.2}-0.01)}\\d=551 ft

Now the perception time is 2.5 second, 20/20 vision  person can read 6 inch letters from 60 x 6 ft.

For 20/40 vision person can read 6 inch letters from 30 x 6 ft=180 ft.

SSD is d=\frac{55 \times 5280 \times 2.5}{60 \times 60}\\d=202 ft

So the minimum distance is given as

Minimum distance=551+202-180 ft\\Minimum distance=573 ft\\

So the sign board must be placed 573 ft ahead of the exit.

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Answer:

Viscosity is notated using the common classification “XW-XX”. The number preceding the “W” (winter) rates the oil's flow (viscosity) at zero degrees Fahrenheit (-17.8 degrees Celsius). The lower the number, the less the oil thickens in cold weather.

7 0
2 years ago
An alternating current E(t) =120 sin(12t) has been running through a simple circuit for a long time. The circuit has an inductan
german

Answer:

Explanation:

we have given E(t)=120 sin(12t)

R=5 ohm

L=0.2 H

ω=12 ( from expression of E)

X_L=0.2\times 12=2.4 ohm

X_C=\frac{1}{\omega \times C}=\frac{1}{12\times 0.043}=1.9379\ ohm

Z=\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C} \right )^2}

Z=\sqrt{5^2+\left ( \2.4-1.9379 )^2}

=5.021 ohm

so amplitude of current =  \frac{v}{z}=\frac{120}{5.021}=23.89

4 0
3 years ago
0/5 pts
Brilliant_brown [7]

Explanation:

150 divide by 150 and that how you do the is you what to divide together 15/ 150 you welcome have a good day is you need something else

4 0
3 years ago
List two ways you can make an informal survey
solniwko [45]

The two ways you can use to make an informal survey are:

  • make field observations
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<h3>What are informal surveys?</h3>

In informal surveys can be regarded as a type of survey that can be made by the researcher by going to the field themselves and this can be done by using different methods or ways.

For instance, the researcher can go out to interview people that can give the data that is needed about the research such as informally asking them questions,  unstructured techniques can also be used to solve critical issues.

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8 0
2 years ago
Read 2 more answers
Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo
AleksandrR [38]

Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

     = ((116.24 - 72.26)/72.26)×100 = 0.6086 ×100 = 60.86%

Therefore, the total valve perimeter is shown to have increased by 60.8%

2) Formula for flow Area (A) = P × L = πkd²

Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

3) The additional benefits of using are:

a) For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

           

7 0
3 years ago
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