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defon
3 years ago
9

On a average work day more than work place firs are reorted​

Engineering
1 answer:
Basile [38]3 years ago
8 0
What is the question? It looks like a statement...
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An undeformed specimen of some alloy has an average grain diameter of 0.040 mm. You are asked to reduce its average grain diamet
Ilia_Sergeevich [38]

Answer:

Yes it is possible. Grain size is often determined by the rate so Firstly reduce the average grain diameter of an undeformed alloy specimen form 0.040 mm to 0.10mm

7 0
2 years ago
1. Which of these is NOT a common type of Camber/Caster adjuster?
Anna35 [415]

Answer: I'm pretty sure Slots is the Answer

8 0
3 years ago
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(Blank) welding involves manual welding with equipment anomalously controls one or more of the windy conditions while (blank) we
shepuryov [24]

Answer:

D.

Explanation:

In automated welding, defined as “welding with equipment that requires only occasional or no observation of the weld, and no manual adjustment of the equipment controls,” the welder's involvement is limited to activating the machine to initiate the welding cycle and observing the weld on an intermittent basis.

8 0
3 years ago
Select four examples of fluid or pneumatic power systems.
Dennis_Churaev [7]

Answer:

Students learn about the fundamental concepts important to fluid power, which includes both pneumatic (gas) and hydraulic (liquid) systems. Both systems contain four basic components: reservoir/receiver, pump/compressor, valve, cylinder.

Explanation:

8 0
3 years ago
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A 200-gr (7000 gr = 1 lb) bullet goes from rest to 3300 ft/s in 0.0011 s. Determine the magnitude of the impulse imparted to the
OlgaM077 [116]

Answer:

The force acting on the bullet F = 84000 \frac{lb ft}{s^{2} }

The value of impulse on the bullet in the given time interval P = 92.4 \frac{lb ft}{sec}

Explanation:

Mass of the bullet ( m ) = 200 gr = 0.028 lb

Initial velocity ( U ) = 0

Final Velocity ( V ) = 3300 \frac{ft}{sec}

Force acting on the bullet F = \frac{m ( V - U )}{t}

⇒ F = \frac{ 0.028 ( 3300 - 0 )}{0.0011}

⇒ F = 84000 \frac{lb ft}{s^{2} }

This is the force acting on the bullet.

Magnitude of the impulse imparted on the bullet  P = F dt  -------- (1)

Put the value of F & dt in above equation we get,

P = 84000 × 0.0011

P = 92.4 \frac{lb ft}{sec}

This is the value of impulse on the bullet in the given time interval.

3 0
3 years ago
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