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3 years ago

Which of the following is an example of seeking accreditation?

Engineering

1 answer:

USPshnik [31]3 years ago

3 0

Answer:

withdraw

blow

view

reviede

draw

throw

show

interview

allow

grow

know

borrow

narrow

renew

flow

bow

swallow

slow

Explanation:

withdraw

blow

view

review

draw

throw

show

interview

allow

grow

know

borrow

narrow

renew

flow

bow

swallow

slow

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Water of dynamic viscosity 1.12E-3 N*s/m2 flows in a pipe of 30 mm diameter. Calculate the largest flowrate for which laminar fl

Naya [18.7K]

Answer:

For water

Flow rate= 0.79128*10^-3 Ns

For Air

Flow rate =1.2717*10^-3 Ns

Explanation:

For the flow rate of water in pipe.

Area of the pipe= πd²/4

Diameter = 30/1000

Diameter= 0.03 m

Area= 3.14*(0.03)²/4

Area= 7.065*10^-4

Flow rate = 7.065*10^-4*1.12E-3

Flow rate= 0.79128*10^-3 Ns

For the flow rate of air in pipe.

Flow rate = 7.065*10^-4*1.8E-5

Flow rate =1.2717*10^-3 Ns

7 0

3 years ago

Phosphorus and nitrogen are included in which category of water pollutants?

Evgesh-ka [11]

Answer: Hello :)

They are in the <u>nutrient pollution</u> category.

Explanation:

3 0

1 year ago

1.The moist unit weights and degrees of saturation of a soil are given: moist unit weight (1) = 16.62 kN/m^3, degree of saturati

alexandr1967 [171]

Answer:

Gs = 2.647

e = 0.7986

Explanation:

We know that moist unit weight of soil is given as

$\gamma_m \ or\ bulk\ density = \frac{(Gs+Se)\times \gamma_w}{(1+e)}$

where, $\gamma_m$ = moist unit weight of the soil

Gs = specific gravity of the soil

S = degree of saturation

e = void ratio

$\gamma_w$ = unit weight of water = 9.81 kN/m3

From data given we know that:

At 50% saturation, $\gamma_m = 16.62 kN/m3$

puttng all value to get Gs value;

$16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}$

Gs - 1.194*e = 1.694 .........(1)

for saturaion 75%, unit weight = 17.71 KN/m3

$17.71 = \frac{(Gs+0.75*e)\times 9.81}{(1+e)}$

Gs - 1.055*e = 1.805 .........(2)

solving both equations (1) and (2), we obtained;

Gs = 2.647

e = 0.7986

6 0

3 years ago

A car is about to start but it blows up. what is the problem with the car<br> ?

ratelena [41]

Answer:

because there is a bomb

6 0

3 years ago

Read 2 more answers

The lift on a spinning circular cylinder in a freestream with a velocity of 30 m/s and at standard sea level conditions is 6 N/m

Evgesh-ka [11]

Answer:

The circulation around the cylinder is 0.163 $\frac{m^{2} }{s}$

Explanation:

Given :

Velocity of spinning cylinder $v = 30$ $\frac{m}{s}$

Sea level density $\rho = 1.23$ $\frac{kg}{m^{3} }$

Sea level span $L = 6$ $\frac{N}{m}$

Lift per unit circulation is given by,

$L = \rho v c$

Where $c =$ circulation around cylinder

$c = \frac{L}{\rho v}$

$c = \frac{6}{1.23 \times 30}$

$c = 0.163$ $\frac{m^{2} }{s}$

Therefore, the circulation around the cylinder is 0.163 $\frac{m^{2} }{s}$

5 0

3 years ago