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MAXImum [283]
3 years ago
7

The phasor technique makes it pretty easy to combine several sinusoidal functions into a single sinusoidal expression without us

ing trigonometric identities. However, you cannot use the phasor technique in all cases. Select the expressions below for which the phasor technique cannot be used to combine the sinusoids into a single expression.
Check all that apply.
O 45 sin(2500t – 50°) + 20 cos(1500t +20°)
O 25 cos(50t + 160°) + 15 cos(50t +70°)
O 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)
O -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
O 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)
Engineering
1 answer:
devlian [24]3 years ago
8 0

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 e^{j(wt+\alpha)}), 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).

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4 0
3 years ago
Calculate the molar heat capacity of a monatomic non-metallic solid at 500K which is characterized by an Einstein temperature of
aleksandr82 [10.1K]

Answer:

Explanation:

Given

Temperature of solid T=500\ K

Einstein Temperature T_E=300\ K

Heat Capacity in the Einstein model is given by

C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}

e^{\frac{3}{5}}=1.822

Substitute the values

C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})

C_v=3R\times \frac{9}{25}\times \frac{1.822}{(0.822)^2}

C_v=0.97\times (3R)            

6 0
3 years ago
A product whose total work content time = 50 min is assembled on a manual production line at a rate of 24 units per hour. From p
valentinak56 [21]

Answer:

a)Cycle time = 2.37 min

b)Numbers of workers =21

c)Stations on the line =24

Explanation:

Given that

Total work content time(TWC) = 50 min

Production rate Rp= 24 units/hr

manning level will be close =1.5

Line balancing efficiency =0.94

a)

Cycle time

T_c=\dfrac{60E}{R_P}

T_c=\dfrac{60\times 0.95}{24}

Cycle time = 2.37 min

b)

Numbers of workers ,W

W=\dfrac{TWC}{T_c}

W=\dfrac{50}{2.37}

W= 21

Numbers of workers =21

c)

Stations on the line(n)

Lets find service time Ts

Ts = Cycle time -  Time for repositioning

Ts = Tc- Tr

Ts= 2.37  - 9/ 60 min

Ts= 2.22 min

We  know that efficiency

\eta=\dfrac{TWC}{n.T_s}

0.94=\dfrac{50}{n\times 2.22}

n=23.94  ⇒n=24

n=24

Stations on the line =24

7 0
3 years ago
A lake is fed by a polluted stream and a sewage outfall. The stream and sewage wastes have a decay rate coefficient (k) of 0.5/d
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Solution :

Given :

k = 0.5 per day

$C_s = 10 \ mg/L \ ; \ \ Q_s= 40 \ m^3/s$

$C_{sw} = 100 \ ppm \ ; \ \ Q_{sw}= 0.5 \ m^3/s$

Volume, V $= 200 \ m^3$

Now, input rate = output rate + KCV ------------- (1)

Input rate  $= Q_s C_s + Q_{sw}C_{sw}$

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                $= 2 \times 10^5 \ mg/s$

The output rate $= Q_m C_{m}$

                          = ( 40 + 0.5 ) x C x 1000

                          $=40.5 \times 10^3 \ C \ mg/s$

Decay rate = KCV

∴$KCV =\frac{0.5/d \times C \  \times 200 \times 1000}{24 \times 3600}$

            = 1.16 C mg/s

Substituting all values in (1)

$2 \times 10^5 = 40.5 \times 10^3 \ C+ 1.16 C$

C = 4.93 mg/L

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3 years ago
Place test tubes containing approximately equal volumes in opposite positions on the rotor. Wait until all positions on the roto
Anvisha [2.4K]

Answer:

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Explanation:

they are giving you step by step instructions to do. i woul hate to have to work with you out anywhere in the world if you have to ask this. i am 15, and i would at least read the d a m n thing before i post it.

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