Answer:
The answer is 4.905 dB
Explanation:
Let say that that signal is sinusoidal i.e Am sin(wt)
Hence the power of the signal
From the question we are given that Amplitude Am = 10mV
substituting this value into the power formula
Power of the signal μW
From the question we where given that the signal to noise ratio is
Note: The dB of a value means the same thing as 10 log of the value
Now to obtain noise power we make it the subject in the above equation
μW
Now to obtain the overall signal gain we multiply the individual gain for the frequency that we are considering i.e 1KHz as our signal
Overall signal gain =
Now that we have gotten this we can now compute the output signal power gain denoted by
W
Now to obtain the overall signal gain we multiply the individual gain for the frequency that we are considering i.e 10KHz as our noise
Output signal to noise ratio (S/N) =
Answer:
0.59
Explanation:
based on Lever's phase rule, we can find the fraction of the α phase in the structure using the following formula:
where
= mass fraction of element B in the entire alloy or mixture = 40%
= mass fraction of element B in the β phase = 90%
= mass fraction of element B in the α phase = 5%
hence we have
∴ The fraction α phase in the structure is ≈ 0.59
Answer:
5.1 Personnel Security. ...
5.2 Physical and Environmental Protection. ...
5.3 Production, Input and Output Controls. ...
5.4 Contingency Planning and Disaster Recovery. ...
5.5 System Configuration Management Controls. ...
5.6 Data Integrity / Validation Controls. ...
5.7 Documentation. ...
5.8 Security Awareness and Training.
Answer:
C. The intensity of 20F signal would be higher, but the lifetime would be unchanged
Explanation:
The half live value is constant, life term of an exponentially decaying quantity, and it a property of the exponential decay equation. Half life does not decay on the initial concentration. Hence, there won't be any change in observation.
C. The intensity of 20F signal would be higher, but the lifetime would be unchanged