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slega [8]
3 years ago
7

A vertical plate is partially submerged in water and has the indicated shape. The height is 3m, the width at the top is 4m, and

the width at the bottom is 8m. Label the shape (trapezoid) appropriately. Express the hydrostatic force against one side of the plate as an integral and evaluate it, if the water level is 1 m from the top of the plate.

Engineering
1 answer:
natka813 [3]3 years ago
7 0

Answer: 1.98 × 10^4 N

Explanation:

Form similar triangle ADE and ABC

a/x= 2/3, a=2/3x

Width of the strip w= 2(4+a) = 8+2a

W= 8 +2 (2/3x)= 8+4/3x

Area of the strip = w Δx

(8 +4/3x) Δx

Pressure on the strip p= pgx= 10^3 ×9.81x= 9810x

But,

Force= Pressure × area= 9810x × (8+4/3x)Δx

Adding the forces and taking lim n to infinity

F total= lim n--> infinity E 9810x × (8+4/3x)Δx

Ftotal= Integral 2,0 9810x × (8+4/3x)Δx

F total= 9810 integral 2, 0 (8+4/3x)dx

= 9810(8+x^2/2 + 4/3x^3/3)2,0

=9810(4x^2 + 4/9x^3)

=9810(4x2^2 + 4/9×2^3-0)

=9810(16 + 32/9)

Hydrostatic force as an integral

Ft= 19.18 ×10^4N

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You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you
telo118 [61]

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

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2 years ago
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Oksana_A [137]

Answer:

The Euler buckling load of a 160-cm-long column will be 1.33 times the Euler buckling load of an equivalent 120-cm-long column.

Explanation:

160 - 120 = 40

120 = 100

40 = X

40 x 100 / 120 = X

4000 / 120 = X

33.333 = X

120 = 100

160 = X

160 x 100 /120 = X

16000 / 120 = X

133.333 = X

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