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Ipatiy [6.2K]
2 years ago
9

X²-12x=0 ПЖ срочно реально решите помагие ​

Engineering
1 answer:
nika2105 [10]2 years ago
4 0

Answer:

формула pq sry i dont speak russian lol but thats the solution of the equation

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2x²-6x+10/x-2 x=2<br><br><br>plsssss<br><br><br><br>​
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5 0
2 years ago
On a average work day more than work place firs are reorted​
Basile [38]
What is the question? It looks like a statement...
8 0
3 years ago
Engineering controls are the physical changes that employers make to the work environment or to equipment that make it safer to
____ [38]

Answer:

Engineering Controls. The best engineering controls to prevent heat-related illness is to make the work environment cooler and to reduce manual workload with mechanization. A variety of engineering controls can reduce workers' exposure to heat: Air conditioning, Increased general ventilation , Cooling fans , Local exhaust ventilation at points of high heat production or moisture, Reflective shields to redirect radiant heat , Insulation of hot surfaces Elimination of steam leaks , Cooled seats or benches for rest breaks , Use of mechanical equipment to reduce manual work, Misting fans that produce a spray of fine water droplets.

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Explanation:

5 0
3 years ago
A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle must be capable of delivering at least 2
Furkat [3]

Answer:

P_{1} = 403,708\,kPa\,(58.553\,psi)

Explanation:

Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}

The initial pressure is:

P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}

The speed at outlet is:

v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}

v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }

v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )

The initial pressure is:

P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}

P_{1} = 403,708\,kPa\,(58.553\,psi)

7 0
3 years ago
Read 2 more answers
I really need help ASAP!!!
ValentinkaMS [17]

Explanation:

He would work on the thing like in the method you work on your question.

8 0
2 years ago
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