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slega [8]
4 years ago
13

What is the flat shape

Engineering
1 answer:
Harlamova29_29 [7]4 years ago
6 0

[C]

The reason why I say this, is because that is the only option that looks like the "flat shape." Notice that the other options have different designs that wouldn't look the same as the flat shape. The flat shape is just a rectangle you piece of cardboard with smaller rectangles at the end of each side. As you can see, shape C shows both smaller rectangles connected without any other design or cur out.

Hope This Helped! Good Luck!

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MATLAB can solve a variety of engineering problems including those requiring simulating differential equations and iterative num
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Answer:

A. True

Explanation:

MATLAB may be defined as a programming platform that is designed specifically for the engineers as well as the scientists to carry out different analysis and researches.

MATLAB makes use of a desktop environment which is tuned for certain iterative analysis and the design processes with a programming language which expresses matrix as well as array mathematics directly.

Thus the answer is true.

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A 2-stage dcv that has an internal pilot does not work well (if at all) on
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Consider a cylindrical nickel wire 1.8 mm in diameter and 2.6 × 104 mm long. Calculate its elongation when a load of 290 N is ap
telo118 [61]

Answer:

e = 3.97*10^-4

Explanation:

1.8 mm = 0.0018 m

2.6*10^4 mm = 26 m

Elongation is The ratio between the stretched length and the original length.

e = L/L0

This is calculated with Hooke's law:

e = σ/E

Where

σ: normal stress

E: elastic constant

σ = P/A

Where

P: normal load

A: cross section

A = π/4 * d^2

Therefore:

e = P / (A * E)

e = 4 * P / (π * d^2 * E)

e = 4 * 290 / (π * 0.0018^2 * 207*10^9) = 3.97*10^-4

8 0
3 years ago
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

6 0
3 years ago
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