Ask question

Ask question

All categories

3 years ago

12

How many trips would one rubber-tired Herrywampus have to make to backfill a space with a geometrical volume of 5400 cubic yard?

The maximum capacity of the machine is 30 cubic yard (heaped), or 40 tons. The material is to be compacted with a shrinkage of 25% (relative to bank measure) and has a swell factor of 20% (relative to bank measure). The material weighs 3,000 lb/cu yd (bank). Assume that the machine carries its maximum load on each trip. Check by both weight and volume limitations

1 answer:

nikklg [1K]3 years ago

8 0

Answer:

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

Explanation:

a) By volume.

The shrinkage factor is:

$\frac{5400cu-yd}{1-0.25} =7200cu-yd$

The volume at loose is:

$V_{loose} =V_{bank} (1+swell-factor)=7200(1+0.2)=8640cu-yd$

If the Herrywampus has a capacity of 30 cubic yard:

$\frac{8640cu-yd}{30cu-yd/trip} =288trip$

b) By weight

The swell factor in terms of percent swell is equal to:

$pounds-per-cubic-yard-loose=\frac{pounds-per-cubic-yard-bank}{\frac{percent-swell}{100}+1 }$

$pounds-per-cubic-yard-loose=\frac{3000}{\frac{20}{100} +1} =2500lb/cu-yd$

The weight of backfill is:

$8640cu-yd*2500\frac{lb}{cu-yd} *\frac{1ton}{2000lb} =10800ton$

The Herrywampus has a capacity of 40 ton:

$\frac{10800}{40ton/trip} =270trip$

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

You might be interested in

Cho P1= XdaN, P2=3.X daN, P= 2.X (daN). a=1m, b=2m,MCN hình tròn d= (100+X)mm1/ Tính nội lực tại các mặt cắt cách ngàm 0,5m; 1m;

DaniilM [7]

Answer:

Please explain it in English so that i can help or you need someone else who can speak Vietnam

5 0

3 years ago

What are the de Broglie frequencies and wavelengths of (a) an electron accelerated to 50 eV (b) a proton accelerated to 100 eV

DaniilM [7]

Answer:

(a) De-Brogie wavelength is 0.173 nm and frequency is 2.42 x 10^16 Hz

(b) De-Brogie wavelength is 2.875 pm and frequency is 4.8 x 10^16 Hz

Explanation:

(a)

First, we need to find velocity of electron. Since, it is accelerated by electric potential. Therefore,

K.E of electron = (1/2)mv² = (50 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 8 x 10^(-18) J

Mass of electron = m = 9.1 x 10^(-31) kg

Therefore,

v² = [8 x 10^(-18) J](2)/(9.1 x 10^(-31) kg)

v = √1.75 x 10^13

v = 4.2 x 10^6 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(9.1 x 10^(-31) kg)(4.2 x 10^6 m/s)

<u>λ = 0.173 x 10^(-9) m = 0.173 nm</u>

The frequency is given as:

Frequency = f = v/λ

f = (4.2 x 10^6 m/s)/(0.173 x 10^(-9) m)

<u>f = 2.42 x 10^16 Hz</u>

(b)

First, we need to find velocity of proton. Since, it is accelerated by electric potential. Therefore,

K.E of proton = (1/2)mv² = (100 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 1.6 x 10^(-17) J

Mass of proton = m = 1.67 x 10^(-27) kg

Therefore,

v² = [1.6 x 10^(-17) J](2)/(1.67 x 10^(-27) kg)

v = √1.916 x 10^10

v = 1.38 x 10^5 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(1.67 x 10^(-27) kg)(1.38 x 10^5 m/s)

<u>λ = 2.875 x 10^(-12) m = 2.875 pm</u>

The frequency is given as:

Frequency = f = v/λ

f = (1.38 x 10^5 m/s)/(2.875 x 10^(-12) m)

<u>f = 4.8 x 10^16 Hz</u>

6 0

3 years ago

Problem 9.11 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain f

BabaBlast [244]

Answer:

the critical flaw length is 10.06 mm

Explanation:

Given the data in the question;

plane strain fracture toughness $K_{tc$ = 92 Mpa√m

yield strength σ $_y$ = 900 Mpa

design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa

Y = 1.15

we know that;

Critical crack length $a_c$ = 1/π( $K_{tc$ / Yσ )²

we substitute

$a_c$ = 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²

$a_c$ = 1/π( 92 Mpa√m / (517.5 Mpa )²

$a_c$ = 1/π( 0.177777 )²

$a_c$ = 1/π( 0.03160466 )

$a_c$ = 0.01006 m = 10.06 mm

Therefore, the critical flaw length is 10.06 mm

{ $a_c$ = ( 10.06 mm ) > 3 mm

The critical flow is subject to detection

5 0

3 years ago

How much calcium chloride will react with100mg of soda ash?

mixer [17]

Answer:

105mg of calcium chloride $CaCl_{2}$

Explanation:

The molecular formula of calcium chloride is $CaCl_{2}$ and the molecular formula of soda ash is $Na_{2}CO_{3}$

First of all you should write the balanced reaction between both compounds, so:

$CaCl_{2}+Na_{2}CO_{3}=2NaCl+CaCO_{3}$

Then you should have the molar mass of the two compounds:

Molar mass of $CaCl_{2}=110.98\frac{g}{mol}$

Molar mass of $Na_{2}CO_{3}=105.98\frac{g}{mol}$

Now with stoichiometry you can find the mass of calcium chloride that reacts with 100mg of soda ash, so:

$100mgNa_{2}CO_{3}*\frac{1molNa_{2}CO_{3}}{105.98gNa_{2}CO_{3}}*\frac{1molCaCl_{2}}{1molNa_{2}CO_{3}}*\frac{110.98gCaCl_{2}}{1molCaCl_{2}}=105mgCaCl_{2}$

6 0

3 years ago

What measures can be taken to mitigate or prevent the formation of stress corrosion cracks?

sasho [114]

Answer:

a) Selection and control of material

b) Control of stress

c) Control of the environment

Explanation:

The measures that can be taken to control and prevent the formation of strss corrosion cracks are:

a) Selection and control of material

It deals with the selection of the best quality material that can resist the stress corrosion cracks to a high extent.

b) Control of stress

It is to control or reduce the intensity of the major stress that causes the stress corrosion cracks.

c) Control of the environment

It includes the prevention of the material from the corrosive environment by removing the component from the region or controlling the environment condition around the object.

7 0

3 years ago