3 years ago

Which of the following describes a polar orbit?

Engineering

1 answer:

Aleksandr-060686 [28]3 years ago

4 0

Low altitude is the answer

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Generally natural shape of stone is in shaped as (a)angular (b)irregular (c)cubical cone shape (d)regular

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Option B. Did i helped?

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A flow of 12 m/s passes through a 6 m wide, 2 m deep rectangular channel with a bed slope of 0. 001. If the mean velocity of flo

prohojiy [21]

Answer:

manning's coefficient is 0.0357

Explanation:

Given:

Velocity of flow, v = 12 m/s

Width of the channel, b = 6 m

Depth of the channel, d = 2 m

bed slope, s = 0.001

mean velocity of flow, V = 1 m/s

now, the velocity is given as:

$V= \frac{1}{n}R^{\frac{2}{3}}S^{\frac{1}{2}}$

where,

n is the manning's coefficient

R is hydraulic mean depth

R = (Area of the channel) / (wetted Perimeter of the channel)

now,

R = (2 × 6) / ((2 × 2) + 6)

R = 12 / 10 = 1.2 m

now, on substituting the values in the equation for velocity, we get

$1= \frac{1}{n}1.2^{\frac{2}{3}}(0.001)^{\frac{1}{2}}$

$n= 1.2^{\frac{2}{3}}(0.001)^{\frac{1}{2}}$

n = 0.0357

hence, the value of manning's coefficient is 0.0357

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A column has a 4.8 cm by 8.7 cm rectangular cross section and a height 4 mm . The column is fixed at both ends and has a lateral

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While recharging a refrigerant system, the charging stops before the required amount of refrigerant has been inserted. What shou

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Explanation:

That is the correct way.

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A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar

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Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading 114.06 m^3/d/m

Explanation:

calculation for single clarifier

$sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d$

$surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2$

$surface area = 314.16 m^2$

volume of tank $V = A\times side\ water\ depth$

$=314.16\times 2 = 628.32m^3$

$Length\ of\ weir = \pi \times diameter of weir$

$= \pi \times 18 = 56.549 m$

overflow rate = $v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2$

Detention time $t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr$

weir loading $= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m$

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3 years ago