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vesna_86 [32]
2 years ago
8

Split the timing circuit below into 3 blocks by drawing 2 vertical lines across it, label each of the blocks stating whether it

is the input, output or process
also I need this for tmrw (14 june) soooo

Engineering
1 answer:
garik1379 [7]2 years ago
3 0

Answer:

i can Help you but iam using my phone so typing is really hectic. reach me via app on +254743503332

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Find the time-domain sinusoid for the following phasors:_________
sattari [20]

<u>Answer</u>:

a.  r(t) = 6.40 cos (ωt + 38.66°) units

b.  r(t) = 6.40 cos (ωt - 38.66°) units

c.  r(t) = 6.40 cos (ωt - 38.66°) units

d.  r(t) = 6.40 cos (ωt + 38.66°) units

<u>Explanation</u>:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = \sqrt{a^2 + b^2}

Ф = direction = tan⁻¹ (\frac{b}{a})

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{-5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{-5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

3 0
3 years ago
Your uncle has given you a newmonitor for your computer. When you attempt to connect it, you notice that none of the ports on th
juin [17]

Answer:

Go on amazon and look for HDMI -> (VGA/DVI/HDMI) what ever you need adapters.

6 0
3 years ago
You are to assess the biomechanics of a male’s arm using his bicep to hold a 20 kg object in his hand. The upper arm is perpendi
cestrela7 [59]

Answer:

Explanation:

The detailed analysis, with free body diagram and step by step calculations with appropriate substitution is as shown in the attached files.

5 0
3 years ago
If a particle moving in a circular path of radius 5 m has a velocity function v = 4t2 m/s, what is the magnitude of its total ac
rjkz [21]

Answer:

8.62m/s²

Explanation:

the particle is experiencing both translational and circular motion

v=4t²

a_{t}=\frac{dv}{dt}=8t

at t=1s, \frac{dv}{dt}=8(1)=8m/s²

a_{c} = \frac{v^{2} }{r}

at t=1, v= 4(1)² = 4m/s

a_{c}=4²/5

a_{c}=3.2m/s²

∴ magnitude of total acceleration, a

a=\sqrt{a_{t} ^{2} + a_{c} {2} }

a=\sqrt{8^{2} +3.2^{2}  }

a=\sqrt{64+10.24}

a=\sqrt{74.24}

a=8.62m/s²

5 0
3 years ago
Read 2 more answers
What should you consider when choosing a shotgun as a firearm?
AnnyKZ [126]
Make sure you meet any minimum legal requirements for caliber, gauge, or the amount of energy produced by the projectile.
7 0
2 years ago
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