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Advocard [28]
3 years ago
9

A section of a two-lane highway has 12-ft lanes and a design speed of 75 mi/h. The section contains a vertical curve and a horiz

ontal curve. The vertical curve connects a-2.5% grade and a 1.5% grade. The PVT of this vertical curve is at station 36 50. The PC of the horizontal curve is located 294 ft before the PVC of the vertical curve. The horizontal curve hassuperelevation of 8% and central angle of 38 degrees. If the vertical and horizontal curves are designed to meet the minimum required lengths for the given design speed, calculate the stations of the PC and PT.

Engineering
1 answer:
spayn [35]3 years ago
6 0

Answer:

Explanation: see attachment below

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c

Explanation:

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1. A thin-walled cylindrical pressure vessel is capped at the end and is subjected to an internal pressure (p). The inside diame
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Find the Rectangular form of the following phasors?
almond37 [142]

Answer:

The angles are missing in the question.

The angles are :

45,     30,    60,     90,    -34,     -56,      20,     -42,  -65,    -15

P=10, P=5,  P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10,   θ = 45°  rectangular coordinates

x = r cosθ  ,   y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

  = 7.07

y =P sinθ = 10 sin 45°

  = 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos  30° = 4.33

y = 5 sin  30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos  60° = 12.5

y = 25 sin  60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos  90° = 0

y = 54 sin  90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos  (-34°) = 53.88

y = 65 sin  (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos  (-56)° = 53.12

y = 95 sin  (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos  20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos  (-42)° = 5.94

y = 8 sin  (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos  (-65)° = 14.79

y = 35 sin  (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos  (-15)° = 144.88

y = 150 sin  (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

6 0
3 years ago
When Ontario
Marat540 [252]

Answer:

Explanation:

This will be possible when setting them up in summer with a certain quantity of sag, they have already know that the cables won't be able to sag any further because of the heat. During winter, when the cables contract because of the cold weather, the sag will therefore be reduced, but much tension will not be put on the cables.

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Answer:

#include <stdio.h>

void SplitIntoTensOnes(int* tensDigit, int* onesDigit, int DecVal){

  *tensDigit = (DecVal / 10) % 10;

  *onesDigit = DecVal % 10;

  return;

}

int main(void) {

  int tensPlace = 0;

  int onesPlace = 0;

  int userInt = 0;

  userInt = 41;

  SplitIntoTensOnes(&tensPlace, &onesPlace, userInt);

  printf("tensPlace = %d, onesPlace = %d\n", tensPlace, onesPlace);

  return 0;

}

4 0
3 years ago
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