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Pie
3 years ago
8

SMAW and GMAW use constant current arc welding machines. True False

Engineering
1 answer:
Oksi-84 [34.3K]3 years ago
4 0
It is true. Good luck
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If a person runs a distance of 0.7 km in 3 min, what is his average speed in kilometres/hour ​
SVEN [57.7K]

Answer:

14 km/hour

Explanation:

8 0
3 years ago
?Why the efficiency of Class A amplifier is very poor​
horsena [70]
Sorry not sure about that
6 0
3 years ago
A cantilever beam AB of length L has fixed support at A and spring support at B.
atroni [7]

The force in the spring will be F =\dfrac{KPl^3}{3EI}.

The deflection of the beam will be \rho = 0.15(\dfrac{KPL^3}{3EI})

<h3>What is a cantilever beam?</h3>

A rigid, horizontally extending structural member known as a cantilever is supported at only one end. Typically, it extends from a solidly affixed flat vertical surface, such as a wall.

Given that:-

  • A cantilever beam AB of length L has fixed support at A and spring support at B.
  • The spring behaves in a linearly elastic manner with stiffness k. If a concentrated load P is applied at B.

The spring force will be calculated as:-

F = kx

Deflection will be given by:-

x = \dfrac{PL^3}{3EI}

The spring force will be calculated by:-

F = \dfrac{KPL^3}{3EI}

The deflection of the beam will be given as:-

\rho = \dfrac{0.15KPL^3}{3EI}

Therefore the force in the spring will be F =\dfrac{KPl^3}{3EI}..The deflection of the beam will be \rho = 0.15(\dfrac{KPL^3}{3EI})

To know more about Cantilever beam follow

brainly.com/question/16791806

#SPJ1

7 0
2 years ago
Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−
sergeinik [125]

Answer:

Power output, P_{out} = 178.56 kW

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, T = 350^{\circ}C

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, \dot{m} = 0.1 kg.s^{- 1}

Diameter of exhaust pipe, d_{h} = 15 cm = 0.15 m

Pressure at exhaust, P' = 50 kPa

temperature, T' =  100^{\circ}C

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

\dot{m} = \frac{Av_{i}}{V_{1}}

0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}

0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}

v_{i} = 3.986 m/s

Now, eqn for compressible fluid:

\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}

Now,

\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}

\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}

\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}

v_{e} = 19.33 m/s

Now, the power output can be calculated from the energy balance eqn:

P_{out} = -\dot{m}W_{s}

P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}

P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW

4 0
3 years ago
The parts of a feature control frame are the tolerance value, the datum references, and the
Elan Coil [88]

Answer:

d

Explanation:

4 0
3 years ago
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