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2 years ago

Find the error in the following pseudo code

Engineering

1 answer:

IRISSAK [1]2 years ago

8 0

Pseudocodes are used as a prototype of an actual program.

The error in the pseudocode is that, the while loop in the pseudocode will run endlessly.

From the pseudocode, the first line is:

<em>Declare Boolean finished = false</em>

The while loop is created to keep running as long as <em>finished = false.</em>

So, for the while loop to end, the finished variable must be updated to true.

This action is not implemented in the pseudocode.

Hence, the error in the pseudocode is that, the while loop is an endless loop

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A horizontal curve on a two-lane highway (10-ft lanes) is designed for 50 mi/h with a 6% superelevation. The central angle of th

fenix001 [56]

Answer:

The PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

Explanation:

From table 3.5 of Traffic Engineering by Mannering

R_v=835

R=835+(10ft/2)= 840 ft.

Now T is given as

T=R tan(Δ/2)

Here Δ is the central angle of curve given as 35°

T=R tan(Δ/2)

T=840 x tan(35/2)

T=840 x tan(17.5)

T=264.85

Now

STA PC=482+72-(2+64.85)=480+07.15

Also L is given as

L=(π/180)RΔ

Here R is the radius calculated as 840 ft, Δ is the angle given as 35°.

L=(π/180)RΔ

L=(π/180)x840 x35

L=512.87 ft

STA PT=480+07.15+5+12.87=485+20.02

Now Ms is the minimum distance which is given as

$M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\$

Here R_v is given as 835

SSD for 50 mi/hr is given as 425 ft from table 3.1 of Traffic Engineering by Mannering

So Ms is

$M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\M_s=835(1-cos(\frac{90 \times 425}{\pi 835}))\\M_s=26.92 ft$

Now for the clearance from the inside lane

Ms=Ms-lane length

Ms=26.92-5= 21.92 ft.

So the PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

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3 years ago

One kilogram of air as an ideal gas executes a Carnot power cycle having a thermal efficiency of 50%. The heat transfer to the a

Alexeev081 [22]

Answer:

(a) the maximum and minimum temperatures for the cycle in K;

Maximum temprature = T1 = 600.05K

Minimum temprature = 300.03K

(b) the pressure and volume at the beginning of the isothermal expansion in bar and m3 respectively;

Pressure (P4) = 0.67842 bar

Volume (V4) = 1.2693 $m^{3}$

(c) the work and heat transfer for each of the four processes in kJ.

$W_{4-1}$ = -215.25 kJ

The answers are clearly explained in the below mentioned document.

Explanation:

Kindly download the below mentioned document, I have explained in quite detail in it. I hope it will help. Thanks.

Download doc

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