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3 years ago

What can be used to relieve stress in a weld.

Engineering

2 answers:

Mekhanik [1.2K]3 years ago

7 0

Answer:

Streamline Stress Relief. "Over the years welders have perfected techniques to relieve stress and minimize distortion: preheating in an oven or with a torch, using heat blankets, and when necessary sending parts to an oven for postweld heat treatment.

Explanation:

Please mark me brainliest

satela [25.4K]3 years ago

3 0

Yes , of course you can

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Compute L, T, M, LC, and R and stations of the BC and EC for the circular curve with the given data of: I (delta) = 22°15′00" an

Mars2501 [29]

Answer:

L = 475.718

T = 240.89 ft

M = 23.0195

LC = 472.728

R = 1225 ft

Explanation:

See the attached file for the calculation.

8 0

3 years ago

If it is desired to lay off a distance of 10,000' with a total error of no more than Â± 0.30 ft. If a 100' tape is used and the

Ira Lisetskai [31]

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

3 0

3 years ago

What 2 forces move the secondary piston ahead?

jekas [21]

Answer:

The primary piston activates one of the two subsystems. The hydraulic pressure created, and the force of the primary piston spring, moves the secondary piston forward.

5 0

3 years ago

Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o

jekas [21]

Complete Question

Complete Question is attached below.

Answer:

$V'=5m/s$

Explanation:

From the question we are told that:

Diameter $d=0.10m$

Power $P=4.0kW$

Head loss $\mu=10m$

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu$

$\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10$

$H_m=(\frac{200*10^3}{1000*9.8}-10)$

$H_m=10.39m$

Generally the equation for Power is mathematically given by

$P=\rho gQH_m$

Therefore

$Q=\frac{P}{\rho g H_m}$

$Q=\frac{4*10^4}{1000*9.81*10.9}$

$Q=0.03935m^3/sec$

Since

$Q=AV'$

Where

$A=\pi r^2\\A=3.142 (0.05)^2$

$A=7.85*10^{-3}$

Therefore

$V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}$

$V'=5m/s$

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3 years ago

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Dafna1 [17]

Answer:

no idea

Explanation:

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3 years ago