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3 years ago

Please answwr the above question screenshot.

Engineering

1 answer:

weeeeeb [17]3 years ago

7 0

Answer:

I have no signal to see your photo

Explanation:

i cant answer that sorry

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True power can only be measured across what?

loris [4]

Answer: mets

Explanation: meets are good

6 0

3 years ago

Read 2 more answers

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

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an el

Allisa [31]

Answer:

theoretical fracture strength = 16919.98 MPa

Explanation:

given data

Length (L) = 0.28 mm = 0.28 × 10⁻³ m

radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m

Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa

solution

we get here theoretical fracture strength s that is express as

theoretical fracture strength = $s_{0} \times \sqrt{\frac{L}{r} }$ .............................1

put here value and we get

theoretical fracture strength = $1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }$

theoretical fracture strength = $16919.98 \times 10^6$

theoretical fracture strength = 16919.98 MPa

3 0

3 years ago

A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show

aleksandrvk [35]

Answer:

$GM$

So the bouy does not float with its axis vertical

Explanation:

From the question we are told that:

Diameter $d=2m$

Length $l=2.5m$

Weight $W=22kN$

Specific weight of sea water $\mu= 10.25kN/m^3$

Generally the equation for weight of cylinder is mathematically given by

Weight of cylinder = buoyancy Force

$W=(pwg)Vd$

Where

$V_d=\pi/4(d)^2y$

Therefore

$22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m$

Therefore

Center of Bouyance B

$B=\frac{y}{2}=0.26m\\\\B=0.75$

Center of Gravity

$G=\frac{I.B}{2}=2.6m$

Generally the equation for\BM is mathematically given by

$BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\$

Therefore

$BG=2.6-0.476\\\\BG=0.64m$

Therefore

$GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\$

Therefore

$GM$

So the bouy does not float with its axis vertical

4 0

3 years ago

4. From two permeability tests it is found that the void ratio and hydraulic conductivity of a normally consolidated clay are 1.

vesna_86 [32]

Answer:

Explanation:

Check attachment for solution

8 0

3 years ago

Please answwr the above question screenshot.​

Please answwr the above question screenshot.