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mr_godi [17]
2 years ago
9

The _______ is a function that describes how the pinna, ear canal, head, and torso change the intensity of sounds with different

frequencies that arrive at each ear from different locations in space. A.combination function B.directional transfer function C.localization function D.inverse-square law E.azimuth
Engineering
1 answer:
scoundrel [369]2 years ago
8 0

Answer:

B: Directional Transfer Function

Explanation:

The function that describes how the pinna, ear canal, head, and torso change the intensity of sounds with different frequencies that arrive at each ear from different locations in space is called Directional Transfer Function.

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What are the following formulas?
4 0
3 years ago
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Purely resistive loads of 24kW, 18kW and 12kW are connected between the neutral and the red, yellow and the blue respectively of
seraphim [82]

Answer:

The phase current in each line conductor are;

I_{R} = 100.17 < 0A

I_{Y} = 75.13< - 120A

I_{B} = 50.08

Explanation:

Given the following data;

Red phase = 24kW,

Yellow phase = 18kW

Blue phase = 12kW

Line voltage = 415V

For a star connected system, we have;

Phase voltage (V_{p} ) = \frac{Line voltage}{\sqrt{3}}

Phase voltage (V_{p} ) = \frac{415}{\sqrt{3}}

Phase voltage (V_{p} ) = 239.6V

The phase sequence for RYB is given by;

V_{R} = 239.6

Phase current (I) = \frac{Phase power}{Phase voltage}

Hence, I = \frac{P}{V}

<em>For the Red phase;</em>

I_{R} = \frac{24000}{239.6

I_{R} = 100.17 < 0A

<em>For the Yellow phase;</em>

I_{Y} = \frac{18000}{239.6

I_{Y} = 75.13< - 120A

<em>For the Blue phase;</em>

I_{B} = \frac{12000}{239.6

I_{B} = 50.08

For the line neutral;

I_{N} =\sqrt{ (I_{R}^{2} +I_{Y}^{2}+I_{B}^{2}-I_{R}I_{Y}-I_{Y}I_{B}-I_{R}I_{B}

Substituting we have, I_{N} = 43.29A

6 0
3 years ago
(a) For the solidification of iron, calculate the critical radius r* and the activation free energy ΔG* if nucleation is homogen
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Answer:

r = 1.46 *10^{-9} m

\Delta G^* = 1.828 *10^{-18} j

Explanation:

given data:

latent heat of fusion \Delta H_f = -1.85*10^9 j/m2

surface free energy = 0.204 j/m2

meltinf point = 1538 degree celcius

\Delta T_c = 273 K

critical radius is given as

r^* = \frac{-2rT_m}{\Delta H_f} *\frac{1}{\DeltaT_c}

= \frac{-2*0.204*(1538+273)}{-1.85*10^9} *\frac{1}{273}

r = 1.46 *10^{-9} m

activation free energy is given as

\Delta G^* = \frac{16\pi r^3 t_m^2}{3\Delta H^2_f} * \frac{1}{\Delta T^2_C}

       = \frac{16\pi 0.204^3*(1538+273)^2}{3*(-1.85*10^9)^2} * \frac{1}{273^2}

\Delta G^* = 1.828 *10^{-18} j

5 0
3 years ago
Looking back on class assignments is a good place to find clues about what to study for upcoming tests.
kondaur [170]

Answer:

True. Memorizing the assignment iteslf well help you study for a test

Explanation:

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3 years ago
The cylindrical tank AB has an 8-in. inner diameter and a 0.32-in. wall thickness. It is fitted with a collar by which a 9-kip f
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Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

4 0
3 years ago
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