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3 years ago

9

The minimum waste compaction ratio reported by the manufacturer of a high-pressure compaction machine is 8:1. What is the corres

ponding percent volume reduction of the waste

1 answer:

Crank3 years ago

6 0

Answer:

12.5%

Explanation:

Compaction ratio= Volume before reduction/volume after reduction

Compaction ratio= 8/1

% reduction in volume= Volume after reduction/Volume before reduction× 100= 1/compaction ratio × 100

% reduction in volume= 1/(8/1) × 100

=12.5

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The speed of sound is 1150 ft/s convert to mile/h

Sonbull [250]

Answer:

784.090909mph

Explanation:

1ft/s=0.681818 mph

1150ft/s=0.681818 x 1150=784.090909

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3 years ago

Ammonia is one of the chemical constituents of industrial waste that must be removed in a treatment plant before the waste can s

Yanka [14]

Answer:

Following is attached the solution or the question given.

I hope it will help you a lot!

Explanation:

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3 years ago

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a)

Sonbull [250]

Answer:

Explanation:

(a)

$\frac{354 mg \, 45 km}{0.0356 kN} = 354 mg \times \frac{1 kg}{10^6 mg} \times 45 km \times \frac{10^3m}{1 km} \times \frac{1}{0.0356 kN} \times \frac{1 kN}{10^3 N} = 0.447 \frac{kg \, m}{N}$

(b)

$0.00453 Mg \times 201 ms = 0.00453 Mg \times \frac{10^3 kg}{1 Mg} \times 201 ms \times \frac{1 s}{10^3 ms} = 0.911 kg \, s$

(c)

$\frac{435 MN}{23.2 mm} = 435 MN \times \frac{10^6 N}{1 MN} \times \frac{1}{23.2 mm} \times \frac{10^3 mm}{1 m} = 18.75 \times 10^9 \frac{N}{m} = 18.75 \frac{GN}{m}$

7 0

4 years ago

A 12-ft high retaining wall has backfill of granular soil with an internal angle of friction of 30 and unit weight of 125 pef. W

irakobra [83]

Answer:

P_p = 27000 psf

Explanation:

given,

height of the retaining wall = h = 12 ft

internal angle of friction (∅)= 30°

unit weight = 125 pcf

Rankine passive earth pressure = ?

k_p is the coefficient of passive earth pressure

$k_p = \dfrac{1 + sin\phi}{1 - sin\phi}$

$k_p = \dfrac{1 + sin30^0}{1 - sin30^0}$

k_p = 3

Passive earth pressure

$P_p = \dfrac{1}{2}k_p \gamma H^2$

$P_p = \dfrac{1}{2}\times 3\times 125 \times 12^2$

P_p = 27000 psf

Rankine passive earth pressure on the wall is equal to P_p = 27000 psf

7 0

3 years ago

The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.

zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

shaft $J\tau$ = $\dfrac{\pi}{2}r^4$

where ;

r = 1 in /2

r = 0.5 in

shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{C/D} = \dfrac{21600}{1079980}$

$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

3 0

3 years ago