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2 years ago

What are the nine Historical periods?

Engineering

1 answer:

Rufina [12.5K]2 years ago

7 0

Answer:

https://quizlet.com/148993376/the-nine-distinct-periods-of-time-flash-cards/

Explanation:

you can find them all : )

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2. A trapezoidal channel has a bottom width of 4 m and side slopes of 2:1 (H:V). If the flow rate is 50 m3/s at a depth of 3.6 m

Daniel [21]

Answer:

Explanation:

Sum of the side slope = 2 + 1 = 3

Length of first slope = 2/3 X 3.6 = 2 X 1.2 = 2.4m

Lenght of second slope = 1/3 X 3.6 = 1.2m

Area of the trapezoidal channel = (2.4 + 1.2)/2 X 3.6 = 1.8 X 3.6 = 6.48m²

Alternate dept = 50m³/6.48m²= 7.716m

4 0

2 years ago

the oscillation of rod oa about o is defined by the relation θ= (3/pi)*(sin*pi*t), where theta and t are expressed in radians an

Mashcka [7]

Answer:

(a) The velocity of the collar = 1.624er-15.54eo in/s

(b) the acceleration of the collar=-49.94er-9.74eo in/s²

(c)the acceleration of collative rod =-3.284er in/s²

Explanation:

Check attachment for calculation

4 0

2 years ago

Suppose you wanted to convert an AC voltage to DC. Your AC voltage source has Vrms= 60V. If you use a full wave rectifier direct

vovangra [49]

Answer:

$V_{dc}=84.15\ V$

Explanation:

Given that

Vrms= 60 V

Vf= 0.7 V

We know that peak value of AC voltage given as

$V_{o}=\sqrt{2}\ V_{rms}$

Now by putting the values

$V_{o}=60\sqrt{2}\ V$

The output voltage of the DC current given as

$V_{dc}=V_{o}-V_f$

$V_{dc}=60\sqrt{2}-0.7\ V$

$V_{dc}=84.15\ V$

Therefore output voltage of the DC current is 84.15 V.

7 0

2 years ago

At an operating frequency of 5 GHz, a 50 lossless coaxial line with insulating material having a relative permittivity r = 2.25

lapo4ka [179]

Answer:

The answer is "150 $\Omega$ ".

Explanation:

Its line length must be converted into wavelengths for the Smith chart to be used.

$\to \beta = \frac{2 \pi }{\lambda } \\\\\to u_p = \frac{\omega }{\beta}\\\\\lambda =\frac{2 pi}{\beta} =\frac{2 \pi U_p}{\omega}=\frac{c}{\sqrt{\varepsilon_r f}}$

$= \frac{3 \times 10^8}{ 2.25 \times (5 \times 10^9)}\\\\= \frac{3}{ 2.25 \times 5\times 10 }\\\\= \frac{3}{ 22.5 \times 5 }\\\\= \frac{3}{ 112.5}\\\\= 0.04 \ m.$

$l =\frac{0.30 }{0.04 } \times \lambda$

$= 7.5 \lambda$

Because it is an integrated half-wavelength amount,

$Z_{in} = Z_L = 150 \ \Omega$

8 0

2 years ago

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas

Lilit [14]

Answer:

A) 209.12 GPa

B) 105.41 GPa

Explanation:

We are given;

Modulus of elasticity of the metal; E_m = 67 GPa

Modulus of elasticity of the oxide; E_f = 390 GPa

Composition of oxide particles; V_f = 44% = 0.44

A) Formula for upper bound modulus of elasticity is given as;

E = E_m(1 - V_f) + (E_f × V_f)

Plugging in the relevant values gives;

E = (67(1 - 0.44)) + (390 × 0.44)

E = 209.12 GPa

B) Formula for upper bound modulus of elasticity is given as;

E = 1/[(V_f/E_f) + (1 - V_f)/E_m]

Plugging in the relevant values;

E = 1/((0.44/390) + ((1 - 0.44)/67))

E = 105.41 GPa

4 0

2 years ago