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VMariaS [17]
2 years ago
3

A triangular plate with base 6 m and height 2 m is submerged vertically in water such that the highest vertex of the plate is 4

meters below the surface and the base is horizontal to the surface.4 m6 m2 mExpress the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s2 for the acceleration due to gravity. Recall that the weight density of water is 1,000 kg/m3.)

Physics
1 answer:
ozzi2 years ago
6 0

Answer:

313.6 kN

Explanation:

Given,

base = 6 m

height = 2 m

vertex height = 4 m

the force on the strip at (6-y) depth

df = ρg(6-y) (3 (2-y) dy)

df = 3ρg (12-8y+y^2) dy

Integration

F = 3\rho g\int\limits^2_{y=0} {12-8y+y^2} \, dy

F = 3\rho g (12y-8\frac{y^2}{2}+\frac{y^3}{3})^2_{0}

F = 3ρg (32/3)

F = 32*1000*9.8

F = 313600 N

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In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad
Crank

Answer:

a)   Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c)      Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

        P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

         A₁ v₁ = A₂ v₂

         v₂ = v₁ A₁ / A₂

We replace

         ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

         ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

         v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

           Q = A v

           Q = A₁ v₁

           Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

            A₁ = π r₁

            A₂ = π r₂

We replace

        Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

      r₁ = d₁ / 2 = 1.00 / 2

      r₁ = 0.500 10⁻² m

      r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

      Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

       Q = 1.57 10⁻² √(12 10³/2550)

        Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

        Q = 1.57 10⁻² √(2 12 10³ / (850 3)

         Q = 4.8 10⁻² m³ / s

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An airplane flying at constant air speed from Indianapolis to St. Louis (assume they are directly east-west of each other) in ca
irga5000 [103]

Answer:

Explanation:

Suppose the distance between the two cities is D and the velocity in calm weather is V . The total time taken in two way travel is given by

Total distance / velocity

= 2 D / V

Suppose velocity of wind is v . Then in one way the velocity of airplane will become V + v and in the return trip its velocity will be V - v

Total time taken

= \frac{D}{V+v} +\frac{D}{V-v}

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1-\frac{v^2}{V^2}

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2D / V

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