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2 years ago

What is the difference between speech recognition and speech synthesis?

Engineering

2 answers:

andrew-mc [135]2 years ago

4 0

Answer:

peech synthesis is being used in programs where oral communication is the only means by which information can be received, while speech recognition is facilitating commu- nication between humans and computers, whereby the acoustic voice signals changes in the sequence of words making up a written text.

Explanation:

taurus [48]2 years ago

4 0

Answer:

Speech synthesis is being used in programs where oral communication is the only means by which information can be received, while speech recognition is facilitating commu- nication between humans and computers, whereby the acoustic voice signals changes in the sequence of words making up a written text.

Explanation:

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(HVAC) PLEASE HELP neeed helpp

meriva

Answer

Sump heater

Explanation

3 0

3 years ago

A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twi

Leviafan [203]

Answer:

F₁ = 1500 N

F₂ = 750 N

$F_{e}$ = 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

= 7.5 x 1000 W

= 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 = F₁ - F₂

750 = 2 F₂ - F₂ ( ∵F₁ = 2 F₂ )

∴F₂ = 750 N

Now F₁ = 2 F₂

F₁ = 2 x F₂

F₁ = 2 x 750

F₁ = 1500 N , this is the maximum force.

Therefore we know,

$F_{max}$ = 3 x $F_{e}$

where $F_{e}$ is centrifugal force

$F_{e}$ = $F_{max}$ / 3

= 1500 / 3

= 500 N

8 0

3 years ago

(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th

arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ( $\dot Q$ ), measured in BTU per hour, is represented by this formula:

$\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})$ (1)

Where:

$\epsilon$ - Emissivity, dimensionless.

$A$ - Surface area of the student, measured in square feet.

$\sigma$ - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

$T_{s}$ - Temperature of the student, measured in Rankine.

$T_{b}$ - Temperature of the bus, measured in Rankine.

If we know that $\epsilon = 0.90$ , $A = 16.188\,ft^{2}$ , $\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}$ , $T_{s} = 554.07\,R$ and $T_{b} = 527.67\,R$ , then the heat transfer rate due to electromagnetic radiation is:

$\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]$

$\dot Q = 417.492\,\frac{BTU}{h}$

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

$Q = \dot Q \cdot \Delta t$ (2)

Where $\Delta t$ is the heat transfer time, measured in hours.

If we know that $\dot Q = 417.492\,\frac{BTU}{h}$ and $\Delta t = \frac{1}{3}\,h$ , then the net energy transfer is:

$Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)$

$Q = 139.164\,BTU$

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

7 0

2 years ago

An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°

Fantom [35]

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

$D = 7.5 \times 10^{-11} m^2/s$

for 0.225% carbon concentration following formula is used

$\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})$

where, erf stand for error function

$\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248$

$0.248 = 1 - erf(\frac{x}{2\sqrt{DT}})$

$erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248$

$erf(\frac{x}{2\sqrt{DT}}) = 0.751$

from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

$\frac{x}{2\sqrt{DT}} = 0.815$

$x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)$

$x = 2.39\times 10^{-3} m$

x = 0.002395 mm

8 0

3 years ago

Tech A says that the brake pedal uses leverage to multiply foot pressure. Tech B says that when braking hard while moving

Nikolay [14]

Tech- A is correct

Explanation:

Leverage is defined as using a tool to gain mechanical influence. The measure of the benefit gained depends on what kind of lever is used and how it is utilized.
Leverage is designed in such a way that it can reproduce the force from your leg many times before any force is transferred to brake fluid.
The brake pedal size and the measure of leverage received depends on the overall design of the brake system.
The second-order lever is used in the brake pedal. The brake pedal applies leverage to populate the force employed to the master cylinder. The effort needed to drive a load depends on the corresponding distance of the load and the work from the fulcrum. The proportion of load and work is known as mechanical advantage.

7 0

3 years ago