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3 years ago

Will give brainliest to whoever answers correctly!!!!

Engineering

2 answers:

Lilit [14]3 years ago

6 0

Answer:

○ D.) m∠1 > m∠4

Explanation:

m∠4 is an acute angle, while m∠1 is an obtuse angle, therefore, you have your above answer.

I am joyous to assist you anytime.

frosja888 [35]3 years ago

5 0

Answer:

B) m<1 + m<2 < m<2 + m<3 + m<4

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Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K. The constant B = 3. 56 times 1014 9cm -3 K-3/2) and the band

meriva

This question is incomplete, the complete question is;

Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K.

The constant B = 3.56×10¹⁴ (cm⁻³ K^-3/2) and the bandgap voltage E = 1.42eV.

Answer: the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³

Explanation:

Given that;

T = 300k

B = 3.56×10¹⁴ (cm⁻³ K^-3/2)

Eg = 1.42 eV

we know that, the value of Boltzmann constant k = 8.617×10⁻⁵ eV/K

so to find the ni for gallium arsenide;

ni = B×T^(3/2) e^ ( -Eg/2kT)

we substitute

ni = (3.56×10¹⁴)(300^3/2) e^ ( -1.42 / (2× 8.617×10⁻⁵ 300))

ni = (3.56×10¹⁴)(5196.1524)e^-27.4651

ni = (3.56×10¹⁴)(5196.1524)(1.1805×10⁻¹²)

ni = 2.1837 × 10⁶ cm⁻³

Therefore the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³

4 0

3 years ago

two pints of ethyl ether evaporate over a period of 1,5 hours. what is the air flow necessary to remain at 10% or less of ethyl

lesya692 [45]

The air flow necessary to remain at the lower explosive level is 4515. 04cfm

<h3>How to solve for the rate of air flow</h3>

First we have to find the rate of emission. This is solved as

2pints/1.5 x 1min

= 2/1.5x60

We have the following details

SG = 0.71

LEL = 1.9%

B = 10% = 0.1 a constant

The molecular weight is given as 74.12

Then we would have Q as

403*100*0.2222 / 74.12 * 0.71 * 0.1

= Q = 4515. 04

Hence we can conclude that the air flow necessary to remain at the lower explosive level is 4515. 04cfm

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2 years ago

Water "bubbles up" h2 = 9 in. above the exit of the vertical pipe attached to three horizontal pipe segments. The total length o

8_murik_8 [283]

The pressure needed at point (1) to produce the flow where water "bubbles up above the exit of the vertical pipe attached to three horizontal pipe segments is 0.750 psi.

<h3>What is Bernoulli's equation?</h3>

The Bernoulli's equation for incompressible fluid can be given as,

$P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2$

Here, ρ is density of fluid, g is acceleration due to gravity, (P) is the pressure v is velocity, h is height of elevation and subscript (1 and 2) is used for point 1 and 2.

Water "bubbles up" h2 = 9 in. Above the exit of the vertical pipe attached to three horizontal pipe segments.

The total length of the 1.75-in.- diameter galvanized iron pipe between point (1) and the exit is 21 inches. h3 = 18 in.

Reynolds number is,

$R_e=\dfrac{V_3D}{v}\\R_e=\dfrac{4.01\dfrac{0.75}{12}}{1.21\times10^{-15}}\\R_e=2.07\times10^4$

The formula for the ratio of pressure to radius, when pressure and velocity at point 2 is zero can be given as,

$\dfrac{p_1}{r}=z_2\left(f\dfrac{l}{d}+\sum k_L\right)\dfrac{v^2}{2g}-\dfrac{v_1^2}{2g}$

Here, f=0.039 and ∑K(L)=4.5. Put the values,

$\dfrac{p_1}{r}=\dfrac{7}{12}\left(0.03\dfrac{21}{0.75}+4.5-1\right)\dfrac{4.01^2}{2\times32.2}\\\dfrac{p_1}{r}=1.73\rm\; ft$

Put the value of r we get,

$p_1=62.4\times1.73\\p_1=108\rm\; ib/ft^2\\p_1=0.750\rm\; psi$

The pressure needed at point (1) to produce the flow where water "bubbles up above the exit of the vertical pipe attached to three horizontal pipe segments is 0.750 psi.

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2 years ago

There is a black-box system (i.e we do not know the transfer function of the system). When we fed a step input into the system,

xxMikexx [17]

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3 years ago

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

andriy [413]

Answer:

Uper Bound = 175.5 GPa, Lower Bound = 85.26 GPa

Explanation:

Rule of mixtures equations:

For a two-phase composite, modulus of elasticity upper-bound expression is,

E(c)(u) = E(m) x V(m) + E(p) x V(p)

Here, E(m) is the modulus of elasticity of matrix, E(p) is the modulus of elasticity of patriciate phase, E(c) is the modulus of elasticity of composite, V(m) is the volume fraction of matrix and V(p) is the volume fraction of composite.

For a two-phase composite, modulus of elasticity lower-bound expression is,

E(c)(l) = E(m) x E(p)/V(m) x E(p)+V(p) x E(m)

Consider the expression of rule of mixtures for upper-bound and calculate the modulus of elasticity upper-bound.

E(c)(u) = E(m) x V(m) + E(p) x V(p), (1)

Calculate the volume fraction of matrix.

V(m) + V(p) = 1

Substitute 0.35 for V(p).

V(m) + 0.35 = 1

V(m) = 0.65

From equation (1);

Substitute 60 GPa for E(m), 390GPa for E(p), 0.65 for V(m) and 0.35 for V(p).

E(c)(u )= E(m) x V(m) + E(p) x V(p)

E(c)(u) = (60 × 0.65) + (390 × 0.35)

E(c)(u) = 175.5 GPa

The modulus of elasticity upper-bound is 175.5GPa.

The modulus of elasticity of upper-bound can be calculated using the rule of mixtures expression. Since the sum of volume fraction of matrix and volume fraction of composite is equal to one V(m) + V(p) = 1. Substitute the value of volume fraction of matrix as 0.69 and obtain the volume fraction of matrix.

Consider the expression of rule of mixtures for lower-bound and calculate the modulus of elasticity upper-bound.

E(c)(l) = (E(m) x E(p))/ (V(m) x E(p) + V(p) x E(m))

Substitute 60 GPa for E(m), 390GPa for E(p), 0.65 for V(m) and 0.35 for V(p).

E(c)(l) = 60 × 390/(0.65 × 390) +(0.35 × 60)

E(c)(l) = 23400/274.5

E(c)(l) = 85.26 GPa

The modulus of elasticity lower-bound is 85.26 GPa.

8 0

4 years ago