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3 years ago

7

Technician A says that weld-through primer can be removed from the immediate weld area to improve weld quality. Technician B say

s that weld-through primer can be removed from adjacent areas to the weld site to improve penetration. Who is right

1 answer:

lisabon 2012 [21]3 years ago

4 0

Answer:

83737373777473737373738388383838

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What did the ancient Greeks use simple machines for?

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Native Americans who used spears to hunt were using wedges. In the third century BC, the Greek scientist Archimedes invented a way to lift water, called the Archimedes screw. It was used to water crops and to move water out of ships.

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Why do side airbags say inflated for several seconds

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Aviation ppl onlyyyy ayoooo, thoughts on the A-10 Thunderbolt II?

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3 years ago

Using the data in the photo write the complex waveform expression

UNO [17]

Answer:

1st Harmonic:

$v(t) = 50\cos(2000\pi t)$

3rd Harmonic:

$v(t) = 9\cos(6000\pi t)$

5th Harmonic:

$v(t) = 6\cos(10000\pi t)$

7th Harmonic:

$v(t) = 2\cos(14000\pi t)$

Explanation:

The general form to represent a complex sinusoidal waveform is given by

$v(t) = A\cos(2\pi f t + \phi)$

Where A is the amplitude in volts of the sinusoidal waveform

Where f is the frequency in cycles per second (Hz) of the sinusoidal waveform

Where $\phi$ is the phase angle in radians of the sinusoidal waveform.

1st Harmonic:

We have A = 50, f = 1000 and φ = 0

$v(t) = 50\cos(2\pi 1000 t + 0) \\\\v(t) = 50\cos(2000\pi t)$

3rd Harmonic:

We have A = 9, f = 3000 and φ = 0

$v(t) = 9\cos(2\pi 3000 t + 0) \\\\v(t) = 9\cos(6000\pi t)$

5th Harmonic:

We have A = 6, f = 5000 and φ = 0

$v(t) = 6\cos(2\pi 5000 t + 0) \\\\v(t) = 6\cos(10000\pi t)$

7th Harmonic:

We have A = 2, f = 7000 and φ = 0

$v(t) = 2\cos(2\pi 7000 t + 0) \\\\v(t) = 2\cos(14000\pi t)$

Note: The even-numbered harmonics have 0 amplitude that is why they are not shown here.

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4 years ago

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Air enters a 200 mm diameter adiabatic nozzle at 195 deg C, 500 kPa and 100 m/s. It exits at 85 kPa. If the exit diameter is 158

ahrayia [7]

Answer:

$v_2 = 160.23 m/s$

$T_2 = 475.797 k$

Explanation:

given data:

Diameter = $d_1 = 200mm$

$t_1 =195 degree$

$p_1 =500 kPa$

$v_1 = 100m/s$

$p_2 = 85kPa$

$d_2 = 158mm$

from continuity equation

$A_1v_1 = A_2v_2$

$v_2 = \frac{\frac{\pi}{4}d_1^2 v_1^2}{\frac{\pi}{4}d_2^2}$

$v_2 = \frac{d_2v_1}{d_2^2}$

$v_2 = [\frac{d_1}{d_2}]^2 v_1$

$= [\frac{0.200}{0.158}]^2 \times 100$

$v_2 = 160.23 m/s$

by energy flow equation

$h_1 + \frac{v_1^2}{2} +gz_1 +q =h_2 + \frac{v_2^2}{2} +gz_2 +w$

$z_1 =z_2$ and q =0, w =0 for nozzle

therefore we have

$h_1 -h_2 =\frac{v_1^2}{2} -\frac{v_2^2}{2}$

$dh = \frac{1}{2} (v_1^2 -v_2^2)$

but we know dh = Cp dt

hence our equation become

$Cp(T_2 -T_1) = \frac{1}{2} (v_1^2 -v_2^2)$

$Cp (T_2 -T_1) = 7836.94$

$(T_2 -T_1) = \frac{7836.94}{1.005*10^3}$

$(T_2 -T_1) = 7.797$

$T_2 = 7.797 +468 = 475.797 k$

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3 years ago