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Ksenya-84 [330]

3 years ago

9

10–25. The 45° strain rosette is mounted on the surface of a shell. The following readings are obtained for each gage: ε a = −20

0(10−6), ε b = 300(10−6), and ε c = 250(10−6). Determine the in-plane principal strains.

1 answer:

vazorg [7]3 years ago

7 0

Answer:

The answer is 380.32×10^-6

Refer below for the explanation.

Explanation:

Refer to the picture for brief explanation.

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What is the steady-state value of the output of a system with transfer function G(s)= 6/(12s+3), subject to a unit-step input?

fenix001 [56]

Answer:

At steady state output will be 2

Explanation:

We have given transfer function $G(S)=\frac{6}{12S+3}$

Input is unit step so $X(S)=\frac{1}{S}$

We know that $G(S)=\frac{Y(S)}{X(S)}$ , here $Y(S)$ , is output

So output $Y(S)=G(S)\times X(S)$

$Y(S)=\frac{1}{S}\times \frac{6}{12S+3}$

Taking 12 common from denominator

$Y(S)=\frac{1}{2S(S+\frac{1}{4})}$

Now using partial fraction

$\frac{1}{2S(S+\frac{1}{4})}=\frac{A}{2S}+\frac{B}{(S+\frac{1}{4})}$

$\frac{1}{2S(S+\frac{1}{4})}=\frac{A(S+\frac{1}{4}+2BS)}{2S(S+\frac{1}{4})}$

$AS+\frac{A}{4}+2BS=1$

On comparing coefficient A=4 and B = -2

Putting the values of A and B in Y(S)

$Y(S)=\frac{4}{2S}-\frac{2}{S+\frac{1}{4}}$

Now taking inverse la place

$y(t)=2-2e^{\frac{-t}{4}}$

Steady state means t tends to infinite

So output at steady state = $y(t)=2-2e^{\frac{-\infty}{4}}$

$y(t)=2-0=2$

3 0

3 years ago

Answer the question faster please

Juliette [100K]

Answer:

No

Explanation:

3 0

3 years ago

A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory

Molodets [167]

Answer:

Explanation:

Mean temperature is given by

$T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}$

Tmean = (Ti + T∞)/2

$T_mean = 107.5^{0}$

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

$\mu$ = 221.6 × 10⁻⁷N.s/m²

$\kappa$ = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

= 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/ $\mu$

Substituting for Pv

Re = 4m/(πD $\mu$ )

= (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

= 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

4 0

3 years ago

A 200 W vacuum cleaner is powered by an electric motor whose efficiency is 70%. (Note that the electric motor delivers 200 W of

Goshia [24]

Answer:

The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.

Explanation:

power efficiency of electric motor = 70% = 0.70

The power output of the vacuum cleaner = $P_o$ = 200 W

The power output of the vacuum cleaner = $P_i$

$Efficiency=\frac{P_o}{P_i}$

$0.70=\frac{200 W}{P_i}$

$P_i=\frac{200 W}{0.70}=285.71 W$

The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.

5 0

3 years ago

Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell w

ivanzaharov [21]

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

5 0

3 years ago