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pshichka [43]
3 years ago
13

A rectangular plate with a width of 19 m and a height of 12 m is located 4 m below a water surface. The plate is tilted and make

s a 35° angle with the horizontal. The resultant hydrostatic force acting on the top surface of this plate is _____. Solve this problem using appropriate software.

Engineering
1 answer:
Kruka [31]3 years ago
4 0

Answer:

F = 33,324,295.32N

Explanation:

We will first of all find the height at which the plate is inclined. We use sine rule in this case

SinΘ = Opp/hyp

Opp=h, hyp=19m, Θ=35°

h = 19xSin35 = 10.899m

Therefore height h=10.899+4 = 14.899m

We then Calculate Area of the plate

Area = 12x19 = 228m²

Finally, we use an online software to calculate the Hydrostatic pressure

The result from the online computation is attached.

The pressure is p = 146159.19Pa

But pressure p is

Pressure=Force/Area

Making Force the subject

Hydrostatic Force = Pressure x Area

F= 146159.19 x 228 = 33,324,295.32N

F = 33,324,295.32N

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A well-designed product will increase?​
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Answer:

true

Explanation:

A well designed product will increase in sells and in stock.

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What 2 forces move the secondary piston ahead?
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Answer:

The primary piston activates one of the two subsystems. The hydraulic pressure created, and the force of the primary piston spring, moves the secondary piston forward.

5 0
3 years ago
Determine the period of each of the following discrete-time signals (if a signal is not periodic, denote its period by infinity)
sergiy2304 [10]

Answer:

a) it is periodic

N = (20/3)k = 20 { for K =3}

b) it is Non-Periodic.

N = ∞

c) x(n) is periodic

N = LCM ( 5, 20 )

Explanation:

We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.

then the period of the signal is given as

N = ( 2π/w₀)K

k is least integer for which N is also integer

Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2

now

a) cos(2π(0.15)n)

w₀ = 2π(0.15)

Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3

so, it is periodic

N = (20/3)k = 20 { for K =3}

b) cos(2n);

w₀ = 2

Now, 2π/w₀ = 2π/2) = π

so, it is Non-Periodic.

N = ∞

c)  cos(π0.3n) + cos(π0.4n)

x(n) = x1(n) + x2(n)

x1(n) = cos(π0.3n)

x2(n) =  cos(π0.4n)

so

w₀ = π0.3

2π/w₀  = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3

∴ N1 = 20

AND

w₀ = π0.4

2π/w₀  = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5

∴ N² = 5

so, x(n) is periodic

N = LCM ( 5, 20 )

6 0
3 years ago
For the speed equation along centerline of a diffuser, calculate the fluid acceleration along the diffuser centerline as a funct
Marrrta [24]

Answer:

a = v\cdot \frac{dv}{dx}, v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1  \right)\cdot x \right]^{-1}, \frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1  \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}

Explanation:

Let suppose that fluid is incompressible and diffuser works at steady state. A diffuser reduces velocity at the expense of pressure, which can be modelled by using the Principle of Mass Conservation:

\dot m_{in} - \dot m_{out} = 0

\dot m_{in} = \dot m_{out}

\dot V_{in} = \dot V_{out}

v_{in} \cdot A_{in} = v_{out}\cdot A_{out}

The following relation are found:

\frac{v_{out}}{v_{in}} = \frac{A_{in}}{A_{out}}

The new relationship is determined by means of linear interpolation:

A (x) = A_{in} +\frac{A_{out}-A_{in}}{L}\cdot x

\frac{A(x)}{A_{in}} = 1 + \left(\frac{1}{L}\right)\cdot \left( \frac{A_{out}}{A_{in}}-1\right)\cdot x

After some algebraic manipulation, the following for the velocity as a function of position is obtained hereafter:

\frac{v_{in}}{v(x)} = 1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1\right) \cdot x

v(x) = \frac{v_{in}}{1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1  \right)\cdot x}

v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1  \right)\cdot x \right]^{-1}

The acceleration can be calculated by using the following derivative:

a = v\cdot \frac{dv}{dx}

The derivative of the velocity in terms of position is:

\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1  \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}

The expression for acceleration is derived by replacing each variable and simplifying the resultant formula.

8 0
3 years ago
Read 2 more answers
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