Answer:
total time = 304.21 s
Explanation:
given data
y = 50% = 0.5
n = 1.1
t = 114 s
y = 1 - exp(-kt^n)
solution
first we get here k value by given equation
y = 1 -
...........1
put here value and we get
0.5 = 1 - e^{(-k(114)^{1.1})}
solve it we get
k = 0.003786 = 37.86 ×
so here
y = 1 -
1 - y =
take ln both side
ln(1-y) = -k ×
so
t =
.............2
now we will put the value of y = 87% in equation with k and find out t
t = ![\sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}](https://tex.z-dn.net/?f=%5Csqrt%5B1.1%5D%7B-%5Cfrac%7Bln%281-0.87%29%7D%7B37.86%2A10%5E%7B-4%7D%7D%7D)
total time = 304.21 s
OA bloom is smaller than a bar
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.
Answer:


Explanation:
The acceleration field is obtained by deriving the components in function of the time. That is to say:

Where
and
.
The velocity components at given point are, respectively:


Lastly, the acceleration components are found:


Answer:
a) Vout= 5V
b) Vout= 5V
c) Vbase= 0.6V
d) Vbase= 0.6V
Explanation:
Consider the circuit shown in attachment
a) When Vin is 0V, the base circuit is not turned, so
Ib=0 and Ic=∞ as transistor is not turned on so
Vout =5V
b) When Vin= 5 V,
Ib= (Vin-Vb)/Rb
Ib=(5-0.6)/1000= 0.0044A
Ic= 0.0044×10=0.044A
Vout= 5- 0.044×1000= not real value
Vout= Vce= 5V
c) voltage drop across Vbase= 0.6V
d) Vbase= 0.6V
In all the above cases, the transistor will not be turned on biasing base voltage and resistor values are very high compared to VCC which is 5V in the given circuit