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frutty [35]
4 years ago
10

Some designers suggest that speech recognition should be used in a telephone menu system. This would allow users to interact wit

h the system by speaking instead of pressing buttons on the dial pad. Give two arguments for and three arguments against the proposal.
Engineering
1 answer:
Ghella [55]4 years ago
5 0

The designers suggest that speech recognition allows for spoken interaction or conversation and spoken prompts or commands.

Explanation:

The speech recognition is used when the user has physical impairments and hands are busy, eyes are occupied and the user cannot read.

We can save time with the usage of speech recognition system. The users are knowledgeable based on the actions available.

The problems faced in speech recognition system are in the noisy environment it has bad microphones and the commands should be learned and remembered.

The other obstacle is error correction is time consuming. the speech production has slow space to speech output provides privacy in public spaces. The disadvantage is it contains large amount of information.

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The gas stream from a sulfur burner is composed of 15-mol% SO2, 20-mol% O2, and 65-mol% N2. This gas stream at 1 bar and 480?C e
galben [10]

Answer: the equilibrium conversion is 0.15 or 15mole %

K=1.33

Explanation:

Given the mole percent of the gases to be as follows

SO2(g) =15% = 0.15mol

O2(g) =20% = 0.20mol

N2(g) = 65% = 0.65mol

The equilibrium conversion reaction is given by

2SO2(g) + O2(g) <------> 2SO3(g)

2 1. 2

According to the reaction 2 moles of SO2 is converted to get 2 mole of SO3

Therefore 0.15 mole will produce 0.15 mole of SO3

So at equilibrium:

2SO2(g) + O2(g) ------> 2SO3(g)

0.15. 0.075. 0.15

Now the equilibrium constant K is given by:

K = [SO3]²/[SO2]²[O2]

Since the container is constant for all the reactants we can neglect the container capacity and use the mole as percent concentration.

K = [0.15]²/[0.15]²[0.075]

K = 1/0.075

K= 1.33

6 0
3 years ago
An oil pump is drawing 44 kW of electric power while pumping oil with rho = 860 kg/m^3 at a rate of 0.07 m^3/s. The inlet and ou
liubo4ka [24]

Answer:

mechanical efficiency  = 54%

Explanation:

given data

power We = 44 kW

density = 860 kg/m³

rate of flow = 0.07 m³/s

inlet diameter d1 = 8 cm

outlet diameter d2 = 12 cm

pressure = 500 kPa

efficiency η = 90%

to find out

mechanical efficiency

solution

we get here shaft power is here

shaft power = efficiency × electric power

shaft power = 0.9 × 44 = 39.6

and

now we get DEmech that is

DE(mech)= m g h + \frac{m}{2} (v² - u ²) .....................1

that is we can write as

DEmech = ρVgh +\frac{\rho V}{2} [(\frac{V}{\pi r2^2})^2- (\frac{V}{\pi r2^2})^2 ]

= VDP + \frac{\rho V^3}{2\pi ^2} × (\frac{1}{\pi r2^2} - \frac{1}{\pi r1^2})^4

Plugging in values we now have   DEmech that is

DEmech = (0.07m^3/s × 500kPa) + \frac{860*0.1^3}{2\pi ^2} *( \frac{1}{0.06^4} -\frac{1}{0.04^4} )  

solve we get

DEmech = 21342 W

DEmech = 21.3 kW

mechanical efficiency is

mechanical efficiency = \frac{21.3}{39.6}

mechanical efficiency = 0.54

mechanical efficiency  = 54%

5 0
3 years ago
In a bicycle race, a bicyclist coasts down a hill with a 7 % grade to save energy. The mass of the bicycle and rider is 80 kg, t
olga_2 [115]

Answer: you need 4567 km but also mph Kikuyus

Explanation:

7 0
3 years ago
Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage cha
alexira [117]

Answer:

Δr=20.45 %

Explanation:

Given that

Rake angle α =  15°

coefficient of friction ,μ = 0.15

The friction angle β

tanβ = μ

tanβ = 0.15

β=8.83°

2φ +  β - α  = 90°

φ=Shear angle

2φ + 8.833° - 15° = 90°

φ = 48.08°

Chip thickness r given as

r=\dfrac{tan\phi}{cos\alpha +sin\alpha\ tan\phi}

r=\dfrac{tan48.08^{\circ}}{cos15^{\circ} +sin15^{\circ}\ tan48.08^{\circ}}

r=0.88

New coefficient of friction ,μ'  = 0.3

tanβ' = μ'

tanβ' = 0.3

β'=16.69°

2φ' +  β' - α  = 90°

φ'=Shear angle

2φ' + 16.69° - 25° = 90°

φ' = 49.15°

Chip thickness r' given as

r'=\dfrac{tan\phi'}{cos\alpha +sin\alpha\ tan\phi'}

r'=\dfrac{tan44.15^{\circ}}{cos49.15^{\circ} +sin49.15^{\circ}\ tan44.15^{\circ}}

r'=0.70

Percentage change

\Delta r=\dfrac{r-r'}{r}\times 100

\Delta r=\dfrac{0.88-0.70}{0.88}\times 100

Δr=20.45 %

8 0
3 years ago
Calculate the number of 12 V batteries (capacity 120 Ah) needed to run a 3 kW DC motor that operates in 240 V. How many hours th
Mrac [35]

Answer:

  • 20 batteries
  • 9.6 hours

Explanation:

To obtain 240 V from 12 V batteries they must be connected in series. The number needed is ...

  240/12 = 20 . . . batteries needed

__

The current draw will be ...

  (3000 W)/(240 V) = 12.5 A

Then the time available from the battery stack is ...

  (120 Ah)/(12.5 A) = 9.6 h

The motor can run 9.6 hours from the series connection.

3 0
3 years ago
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