Q1: The first option should always be to get out safely (RUN)

g A 30-m-diameter sedimentation basin has an average water depth of 3.0 m. It is treating 0.3 m3/s wastewater flow. Compute over

stepladder [879]

Answer:

The overflow rate is 4.24×10^-4 m/s.

The detention time is 7069.5 s

Explanation:

Overflow rate is given as volumetric flow rate ÷ area

volumetric flow rate = 0.3 m^3/s

area = πd^2/4 = 3.142×30^2/4 = 706.95 m^2

Overflow rate = 0.3 m^3/s ÷ 706.95 m^2 = 4.24×10^-4 m/s

Detention time = volume ÷ volumetric flow rate

volume = area × depth = 706.95 m^2 × 3 m = 2120.85 m^3

Detention time = 2120.85 m^3 ÷ 0.3 m^3/s = 7069.5 s

6 0

2 years ago

How do you get your drivers lisnes when your 15

prohojiy [21]

Answer:

You take a drivers test!

Explanation:

5 0

1 year ago

The speed of an aircraft is given to be 260 m/s in air. If the speed of sound at that location is 330 m/s, the flight of the air

navik [9.2K]

Answer:

$Ma=\frac{260}{330} \\Ma=0.7878$

So, Ma < 1 Flow is Subsonic

Explanation:

Mach Number:

Mach Number is the ratio of speed of the object to the speed of the sound. It is used to categorize the speed of the object on the basis of mach number as sonic, supersonic and hyper sonic. (It is a unit less quantity)

Mach < 1 Subsonic

Mach > 1 Supersonic

Ma= Speed of the object/Speed of the sound

$Ma=\frac{260}{330} \\Ma=0.7878$

So, Ma < 1 Flow is Subsonic

8 0

3 years 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

2 years ago

A ball thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s. Determine (a) how hig

Masteriza [31]

Answer:

A.) 62.5 ft

B.) 3.58 seconds

C.) 8.58 seconds

Explanation:

A.) Given that a ball is thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s

To determine how high above the top of the building the ball will go before it stops at B, let us use the third equation of motion.

V^2 = U^2 - 2gH

Since the ball is going up, g will be negative. And at maximum height, V = 0

Substitute all the parameters into the formula

0 = 35^2 - 2 × 9.8 × H

19.6H = 1225

H = 1225/19.6

H = 62.5 ft

(B) The time tAB it takes to reach its maximum height will be achieved by using second equation of motion

H = Ut - 1/2gt^2

Substitutes all the parameters into the formula

62.5 = 35t - 1/2 × 9.8 × t^2

62.5 = 35t - 4.9t^2

4.9t^2 - 35t + 62.5 = 0

Let's use quadratic equations to find t

Divide all by 4.9

t^2 - 7.143t + 12.755 = 0

t^2 - 7.143t + 3.57^2 = - 12.755 + 3.57^2

( t - 3.57)^2 = 0.000102

( t - 3.57 ) = +/-( 0.01 )

t = 3.57 + 0.01

t = 3.58 seconds

Ignore the negative one.

(C) the total time tAC needed for it to reach the ground at C from the instant it is released.

When the object is falling back from B, the initial velocity = 0. And the height h will be 60 + 62.5 = 122.5 ft

Using equation 2 of equations of motion again.

h = 1/2gt^2

122.5 = 1/2 × 9.8 × t^2

122.5 = 4.9t^2

t^2 = 122.5/4.9

t^2 = 25

t = 5

Total time = 5 + 3.58 = 8.58 seconds

3 0

2 years ago