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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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4.116 The lid of a roof scuttle weighs 75 lb. It is hinged at corners A and B and maintained in the desired position by a rod CD
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Answer:

(a) The magnitude of force is 116.6 lb, as exerted by the rod CD

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Explanation:

Step by step working is shown in the images attached herewith.

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Answer:

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Explanation:

only if there signal is turned on

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For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho
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Answer:d

Explanation:

Given

Temperature=200^{\circ}\approc 473 K

Also \gamma for air=1.4

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no.=\frac{V}{\sqrt{\gamma RT}}

(a)1000 km/h\approx 277.78 m/s

Mach no.=\frac{277.78}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.63

(b)500 km/h\approx 138.89 m/s

Mach no.=\frac{138.89}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.31

(c)2000 km/h\approx 555.55 m/s

Mach no.=\frac{555.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=1.27

(d)200 km/h\approx 55.55 m/s

Mach no.=\frac{55.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

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