Answer: the scalar magnitude on lead I is 0.5 mV
Explanation:
Given that;
scalar magnitude on lead II = 1 mV
scalar magnitude on lead III = 0.5 mV
the scalar magnitude on lead I = ?
we know that;
Lead I Voltage = LA - RA -----------let this be equation 1
where LA is left arm electrode and RA is right am electrode
Also
Lead II = LL - RA
where LL is the left leg of electrode
we substitute
1 mV = LL - RA ---------------------let this be equation 2
Again
Lead III = LL - LA
we substitute
0.5 mV = LL - LA ------------------let this be equation 3
now subtract equation 3 and 2
1 mV - 0.5 mv = LL - RA - (LL - LA)
0.5 mV = LL - RA - LL + LA
0.5 mV = -RA + LA
0.5 mV = LA - RA
now taking a look at our equation 1 ( Lead I Voltage = LA - RA )
hence, Lead I Voltage = LA - RA = 0.5 mV
Therefore the scalar magnitude on lead I is 0.5 mV
Which system provides an easier way for people to communicate with a computer than a graphical user interface (GUI)?
a) data mining
b) transaction-processing
c) natural language processing
d) data abstraction
Answer:
C.
Explanation:
Natural language processing or NLP can be defined as an artificial intelligence that aids computers to understand, interpret, and manipulate human language. NLP is subfiled of many disciplines, for instance, artifical intelligence, linguistics, computer science, etc. NLP helps computers to interact with humans in their own language.
So, the easier way through which people can communicate with computers apart from GUI is NLP. Thus option C is correct.
If the length of the arm were to increase, how would the required torque change? a. Required torque would increase b. Required torque would decrease c. Required torque would remain the same d. It is impossible to calculate required torque with the values given.
Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distance that a 1 (GHz) wave travels in graphite such that its field intensity is reduced by 30 (dB).
Answer:
the answer is below
Explanation:
a) The conductivity of graphite (σ) is calculated using the formula:
[tex]\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}[/tex]
where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012
Substituting:
[tex]\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m[/tex]
b) f = 1 GHz = 10⁹ Hz.
[tex]\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10} e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10} e=-1.5 \\\\z=\frac{-1.5}{log_{10} e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm[/tex]
A blown fuse or tripped circuit breaker shows
Answer:
Your house is in need of a service upgrade, or it may indicate that your house has too few circuits.
Explanation:
It's a sign that you are making excessive demands on the circuit and need to move some appliances and devices to other circuits.
