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Archy [21]
3 years ago
5

Instruments in airplane A indicate that with respect to the air the plane is headed 30◦ north of east with an airspeed of 300 mi

/h. At the same time radar on ship B indicates that the relative velocity of the plane with respect to the ship is 280 mi/h in the direction 33◦ north of east. Knowing that the ship is steaming due south at 12 mi/h, determine (a) the velocity of the airplane, (b) the wind speed and direction.
Engineering
1 answer:
mamaluj [8]3 years ago
4 0

Answer:

velocity of the plane = 274 miles/hour at 30.9⁰

velocity and direction of the wind = 26.7 miles/hour at 20.8⁰

Explanation:

Let the velocities be resolved using the vector method.

Let the x - axis be directed to the east

similarly, we shall direct the y axis upwards- to the north

Airspeed is given by

300 mil/h. (30⁰) = 300 (cos30⁰i + sin 30⁰j

plane relative to ship

= 12 miles/hour

= 12j

  • velocity of the plane is given by:

v =A_{p}  + Vijb

  = -12 j + 280 (cos 33⁰i + sin 33⁰j)

  = 234.83 i + 140.50 j

V_{a} = 274 miles per hour

  • velocity of the wind is given by the following equation:

v_{w}  = +WIA= A-

    = 234.831 + 140.50J - 300(cos 30⁰i + sin 30⁰j)

   = -24.98i - 9.50 j

   = 26.7 miles per hour at 20.8⁰

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Answer:

A. 288,030.91 cy

B. The amount of water that must be removed from the natural material is 483541.04254 gallons of water

Explanation:

The natural material in the barrow properties are;

The mass unit weight, γ = 110.0 pcf

The water content, w = 6%

The specific gravity of the soil solids, G_s = 2.63

The desired dry unit weight, \gamma _d = 122 pcf

The water content, w₁ = 5.5 %

The net section volume, V_T = 245,000 cy = 6,615,000 ft³

A.  \gamma _d = W_s/V_T

∴ W_s = V_T × \gamma _d = 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs

w = (W_w/W_s) ×  100

∴ W_w = (w/100) × W_s = (6/100) × 807030000 lbs = 48421800 lbs

The weight of solids

W = W_s + W_w = 807030000 lbs + 48421800 lbs = 855451800 lbs

V = W/γ = 855451800 lbs/(110.0 lb/ft.³) = 7776834.54545 ft.³ = 288,030.91 cy

V = 288,030.91 cy

The amount of cubic yards of borrow required = 288,030.91 cy

B. The volume of water in the required soil is found as follows;

W_{w1} = (w₁/100) × W_s = (5.5/100) × 807030000 lbs = 44386650 lbs

The amount of water that must be added =  W_{w1} - W_w = 44386650 lbs - 48421800 lbs = -4,035,150 lbs

Therefore, 4,035,150 lbs of water must be removed

The density of water, ρ = 8.345 lbs/gal

Therefore, V = 4,035,150 lbs/(8.345 lbs/gal) = 483541.04254 gal  of water must be removed from the natural material

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A digital automatic level is used to profile a road centerline. The backsight reading is 3.57 ft on BM #1, which has an elevatio
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Answer:

209.55 ft

Explanation:

Given Data:

Benchmark:

Reduced Level or Elevation = 210.50

Height of Instrument = Reduced Level + Back sight Reading

Height of Instrument = 210.50 + 3.57 = 214.07 ft

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Back sight Reading = 2.91 ft

Fore Sight Reading = 4.52

Reduced Level or Elevation of Turning Point = Height of Instrument – fore sight Reading

Reduced Level or Elevation of Turning Point = 214.07 – 4.52 = 209.55 ft

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Height of Instrument at Turning Point = 209.55 + 2.91 = 212.46 ft

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1. Sewage-treatment plant, a large concrete tank initially contains 440,000 liters liquid and 10,000 kg fine suspended solids. T
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200m³/440m³ = 5/11

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Amount left; = 10000 x (5/11) = 4545 kg

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Concentration = amount left/initial volume

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