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Arte-miy333 [17]

2 years ago

13

Find the velocity and rate of flow of water through a rectangular channel of 6m wide and 3m deep when it's running full. The cha

nnel is having a bed slope of 1 in 2000. Take Chezy's coefficient, C=55

1 answer:

Elza [17]2 years ago

5 0

Answer:

V = 1.5062 m/s

Explanation:

look to the photos

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A shipment of rebar that weighs 745 kg would weigh roughly how much in pounds

Andre45 [30]

Answer:

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Your task is to fill in the missing parts of the C code to get a program equivalent to the generated assembly code. Recall that

Rudik [331]

Answer:

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Explanation:

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6 0

2 years ago

Hot exhaust gases of an internal combustion engine are to be used to produce saturated water vapor at 2 MPa pressure. The exhaus

Anastaziya [24]

Answer:

The flowrate of water is 0.03556kg/s

Explanation:

Exhaust gases inlet temperature T1=4000C

Water inlet temperature T3=150C Exit Pressure of water as saturated vapor P4=2MPa

Mass flow rate of exhaust gases Heat lost to the surroundings Qgases=32kg/min

Mass flow rate of exhaust gases is 15 times that of the water

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2 years ago

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Sharon is designing a house in an area that receives a lot of rainfall all year. Which material should she use to stick the wood

kakasveta [241]

Explanation:

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4 0

2 years ago

The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of forc

Sonbull [250]

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

$\theta$ = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑ $f_{y}$ = 0, ∑M = 0

If ∑ $M_{A}$ = 0

⇒ $F_{BC}$ ×0.9 - P × 0.6 = 0

⇒ $F_{BC}$ ×3 - P × 2 = 0

⇒ $F_{BC}$ = $\frac{2P}{3}$

If ∑ $M_{B}$ = 0

⇒ $F_{AD}$ ×0.9 = P × 0.3

⇒ $F_{AD}$ ×3 = P

⇒ $F_{AD}$ = $\frac{P}{3}$

Now,

Area, A = $\frac{\pi }{4} X (0.012)^{2}$ = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , $\delta$ = $\frac{P l}{A E}$

Now,

$\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 9.1626 × 10⁻⁹ P

$\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 1.83253 × 10⁻⁸ P

Given that,

$\theta$ = 0.015°

⇒ $\theta$ = 2.618 × 10⁻⁴ rad

So,

$\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}$

⇒2.618 × 10⁻⁴ = ( 1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

6 0

2 years ago