A thermistor is a temperature‐sensing element composed of a semiconductor material, which exhibits a large change in resistance
1 answer:
The temperature at which the resistance is is
Explanation:
For the thermistor in this problem, the relationship between temperature and resistance is linear.
We have:
when the temperature is
when the temperature is
Assuming a straight-line relationship, we can find the slope of the line:
Now that we know the slope, we can extrapolate the temperature when the resistance is
In fact, we can write:
And solving for ,
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Answer:
The graphs are attached below
Explanation:
Ans) We know,
F(t) = K(t - to) + zΔ∅ ln[(1 + F(t)/zΔ∅]
where K = hydraulic conductivity =0.1 cm/hr
z = capillary suction =20cm
Δ∅ = ∅ (1 - Se)
∅ = effective porosity , Se = initial moisture content
Δ∅ = 0.40(1 - 0.15)
Δ∅ = 0.34
Now, F(t) = 0.1(t - to) + 20(0.34) ln[ 1 + F(t)/6.8]
F(t) = 0.1(t - to) + 6.8 ln [1+ 0.147 F(t)]
Also, infiltration rate (f),
f = K [(zΔ∅ + F)/F]
f = 0.40 [6.8 + F]/F
Condition of ponding,
Ponding time tp = K zΔ∅ i(i-k)
where, i = rainfall rate (1cm/hr)
tp = 0.40(6.8) / 0.60
tp= 4.53 hours
Now, cumulative infiltraton at ponding Fp = i tp
Fp = 1 x 4.53 or 4.53 cm
For infiltration at time less then ponding time , infiltration rate = rainfall rate
For t = 0.25 hr , f = 1cm/hr ; F = 0.25 x 1 = 0.25 cm
For t = 0.50 hr , f = 1cm/hr ; F = 0.50 cm
For t = 0.75 hr , f = 1cm/hr ; F = 0.75 cm
For t = 1 hr , f = 1cm/hr ; F = 1cm/hr
For t > tp ,
Equivalent time origin(to),
to = tp - 1/K [ Fp - z Δ∅ ln (1+ Fp/zΔ∅)
to = 4.53 - 1/0.40[ 4.53 - 6.8 ln(1 + 0.66)
to = 4.53 - 2.5 ( 4.53 - 3.44)
to = 1.82 hr
Hence,
F = 0.10( t - 1.82) + 6.8 ln[ 1+ 0.147F(t)]
Solving above equation for t by assuming F, and further solving equation for infiltration rate
Ans (a) Following is required curve :
O.S 1-5 2 time Chr)ホー Lt 2 time Chr)
Ans (c) Following is required table :
Ans (d) For time step t = 0.25 hrs
At t < tp ; i = f = 1 cm/hr
F = i x t
F = 1 x 0.25
F = 0.25 cm
Answer:
The phasor technique can't be applied directly in the following cases:
a) 45 sin(2500t – 50°) + 20 cos(1500t +20°)
b) 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)
c) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
d) 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)
Explanation:
For a) and c), it is not possible to use the phasor technique, due this technique is only possible when the sinusoidal signals to be combined are all of the same frequency.
This is due to the vector representing a signal is showed as a fixed vector in the graph( which magnitude is equal to the amplitude of the sinusoid and his angle is the phase angle with respect to cos (ωt)), which is rotating at an angular speed equal to the angular frequency of the sinusoidal signal that represents, like a radius that shows a point rotating in a circular uniform movement.
This rotating vector represents a sinusoidal signal, in the form of a cosine (as the real part of the complex function ), so it is not possible to combine with functions expressed as a sine, even though both have the same frequency.
If we look at the graphs of cos (ωt) and sin (ωt) we can say that the sin lags the cos in 90º, so we can say the following:
sin (ωt) = cos (ωt-90º)
This means that in order to be able to represent a sine function as a cosine, we need to rotate it 90º in the plane clockwise.
This is the reason why before doing this transformation, it is not possible to use the phasor technique for b) and d).