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Ivanshal [37]
4 years ago
6

The hot-wire anemometer is an instrument used for measuring velocities or temperatures. If, during its calibration, the output s

ignal E is measured as 0, 1.7, 3.3, and 5.6 V at velocities V of 0, 1, 1.5, and 2 m/s, obtain the gradient dE/dV at V = 0 m/s, using the polynomial representation of the function E(V).

Engineering
1 answer:
Alexus [3.1K]4 years ago
6 0

Answer:

Explanation:

An expression was assumed for E(V) = av^2 + bv + c

each values for substituted for corresponding values of the velocity and a three unknown equation was formed, which was solved simultaneously to get the three unknowns (a, b and c).

The unknowns was substiutted back ito the original equation.

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xeze [42]

Answer:

I don't have robux

Explanation:

but i love adopt me

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3 years ago
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A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an
Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here  thermodynamic equation that

energy equation that is

h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w

put here value with

turbine is insulated so q = 0

so here

3015.4 *1000 + \frac{10^2}{2}  =  2431.7 * 1000 + \frac{90^2}{2}  + w

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0
3 years ago
Ruler game, HELPPPPP
viktelen [127]
D! :D
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3 years ago
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Low-level coding means that...
svet-max [94.6K]

Answer:

a.) a component item is coded at the lowest level at which it appears in the BOM structure is the correct answer.

Explanation:

  • Low-level coding is a kind of programming language used in BOM structures and it carries basic commands that are identified by a computer.
  • The two types of low-level coding are
  • Assembly language.
  • machine language.
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7 0
4 years ago
The time to half-maximum voltage is how long it takes the capacitor to charge halfway. Based on your experimental results, how l
satela [25.4K]

Answer:

Time taken for the capacitor to charge to 0.75 of its maximum capacity = 2 × (Time take for the capacitor to charge to half of its capacity)

Explanation:

The charging of a capacitor/the build up of its voltage follows an exponential progression and is given by

V(t) = V₀ [1 - e⁻ᵏᵗ]

where k = (1/time constant)

when V(t) = V₀/2

(1/2) = 1 - e⁻ᵏᵗ

e⁻ᵏᵗ = 0.5

In e⁻ᵏᵗ = In 0.5 = - 0.693

-kt = - 0.693

kt = 0.693

t = (0.693/k)

Recall that k = (1/time constant)

Time to charge to half of max voltage = T(1/2)

T(1/2) = 0.693 (Time constant)

when V(t) = 0.75

0.75 = 1 - e⁻ᵏᵗ

e⁻ᵏᵗ = 0.25

In e⁻ᵏᵗ = In 0.25 = -1.386

-kt = - 1.386

kt = 1.386

t = 1.386(time constant) = 2 × 0.693(time constant)

Recall, T(1/2) = 0.693 (Time constant)

t = 2 × T(1/2)

Hope this Helps!!!

3 0
3 years ago
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