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bearhunter [10]
3 years ago
12

Explain the prosses of welding

Engineering
2 answers:
klio [65]3 years ago
7 0

Answer:

Welding is a fabrication process whereby two or more parts are fused together by means of heat, pressure or both forming a join as the parts cool. Welding is usually used on metals and thermoplastics but can also be used on wood.

Explanation:

serious [3.7K]3 years ago
4 0
Welding is like a Fabrication Process ,which is where two or more parts are fused together by heat,Pressure or both forming or joining together as the parts cool.HOPE THIS HELPED!
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What are typical uses of Mainframe computer?​
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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°)

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Read 2 more answers
transformer has 26 turns on the primary coil and 720 turns on the secondary coil. If this transformer is to produce an output of
hram777 [196]
<h2>Corrected Question:</h2>

A step-up transformer has 26 turns on the primary coil and 720 turns on the secondary coil. If this transformer is to produce an output of 4100 V with a 16-mA current, what input current and voltage are needed?

<h2>Answer:</h2>

The input current and voltage needed are 443 mA and 148 V respectively.

<h2>Explanation:</h2>

In a step-up transformer, the relationship between the number of turns in its primary coil (N_{p}), the number of turns in its secondary coil (N_{s}), the input voltage (V_{p}), the output voltage (V_{s}), the input current (I_{p}), and the output current (I_{s}) is given by;

\frac{N_{s} }{N_{p} } = \frac{V_{s} }{V_{p} } = \frac{I_{p} }{I_{s} }

This implies that;

\frac{N_{s} }{N_{p} } = \frac{V_{s} }{V_{p} }  ---------------------(i)

\frac{N_{s} }{N_{p} } = \frac{I_{p} }{I_{s} }  ---------------------(ii)

From the question;

N_{p} = 26 turns

N_{s} = 720 turns

V_{s} = 4100V

I_{s} = 16mA = 16 x 10⁻³A

(a) Substitute these values into equation (ii) as follows;

\frac{720}{26} = \frac{I_{p} }{16*10^{-3}}

Solve for I_{p};

I_{p} = 720 x 16 x 10⁻³ / 26

I_{p} = 443 x 10⁻³

I_{p} = 443 mA

Therefore the input current needed is 443mA

(b) Also, substitute those values into equation (i) as follows;

\frac{720}{26} = \frac{4100 }{V_{p} }

Solve for V_{p};

V_{p} = 4100 x 26 / 720

V_{p} = 148 V

Therefore, the input current needed is 148 V

5 0
3 years ago
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