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spayn [35]
3 years ago
13

From the top of a vertical cliff 80m high, the angles of depression of 2 buoys lying due west of the cliff are 23° and 15° respe

ctively. How far are the buoys apart?​
Engineering
1 answer:
Umnica [9.8K]3 years ago
6 0

Answer:

  110 m

Explanation:

The geometry of the problem can be modeled by a right triangle. The height of the cliff forms one leg, and the distance to a buoy forms the other leg. The angle of depression is opposite the height of the cliff. An appropriate trig relation is ...

  Tan = Opposite/Adjacent

Solving for the Adjacent side (the distance to the buoy), we find ...

  Adjacent = Opposite/Tan

  distance to buoy 1 = (80 m)/tan(23°) ≈ 188.468 m

  distance to buoy 2 = (80 m)/tan(15°) ≈ 298.564 m

Then the distance between the buoys is ...

  298.564 -188.468 m ≈ 110 m

The buoys are about 110 meters apart.

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A metallic material with yield stress of 140 MPa and cross section of 300 mm x 100 mm, is subjected to a tensile force of 8.00 M
Readme [11.4K]

Answer:Yes,266.66 MPa

Explanation:

Given

Yield stress of material =140 MPa

Cross-section of 300\times 100 mm^2

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\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}

\sigma =266.66 \times 10^{6} Pa

\sigma =266.66 MPa

Since induced stress  is greater than Yield stress therefore Plastic deformation occurs

8 0
3 years ago
Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an
ser-zykov [4K]

Answer:

volumetric flow rate = 0.0251 m^3/s

Velocity in pipe section 1 = 6.513m/s

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the  volume will be mass/density

25/997 = 0.0251 m^3/s

volumetric flow rate = 0.0251 m^3/s

Average velocity calculations:

<em>Pipe section A:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s

<em>Pipe section B:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s

7 0
3 years ago
A fluid of specific gravity 0.96 flows steadily in a long, vertical 0.71-in.-diameter pipe with an average velocity of 0.90 ft/s
KengaRu [80]

Answer:

0.00650 Ib s /ft^2

Explanation:

diameter ( D ) = 0.71 inches = 0.0591 ft

velocity = 0.90 ft/s ( V )

fluid specific gravity = 0.96 (62.4 )  ( x )

change in pressure ( P ) = 0 because pressure was constant

viscosity =  (change in p - X sin∅ ) D^{2} / 32 V

              = ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2  / 32 * 0.90

              = - 59.904 sin (-90) * 0.0035 / 28.8

              = 0.1874 / 28.8

  viscosity = 0.00650 Ib s /ft^2

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