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<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 =
Ф = direction = tan⁻¹ ()
(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 =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
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 =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
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 =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
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 =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
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°)
Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress = 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress
Making thickness t the subject; we have
t = 0.0056 m
t = 5.6 mm
For longitudinal stress.
t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm