All categories

2 years ago

Estimate the quantity of soil to be excavated from the borrow pit

Engineering

1 answer:

Serggg [28]2 years ago

6 0

Answer

1056

Explanation:

for example

A soil is to be excavated from a borrow pit which has a density of 1.75g/cc and water content of 12%. The G is 2.7 the soil is compacted to that water content of 18% and dry density of 1.65g/cc. for 1000 m3 of soil used in fill estimate

Quantity of soil to be excavated from pit in m3

You might be interested in

Air flows through a 0.25-m-diameter duct. At the inlet the velocity is 300 m/s, and the stagnation temperature is 90°C. If the M

Naddika [18.5K]

Answer:

a. 318.2k

b. 45.2kj

Explanation:

Heat transfer rate to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

See attachment for detailed analysis

7 0

3 years ago

When an object is moving, we use the following coefficient for friction calculations a)-μk b)-μs c)-γk d)- γs

Reika [66]

Answer: $\mu_{k}$

Explanation:

We use kinetic friction when a body is moving i.e. $\mu_{k}$ for calculations.

Static friction is used when a body is in rest while kinetic friction is used when a body is moving and its value is quite low as compared to static friction .

Static friction value increases as we apply more force while kinetic friction occurs when there is relative motion between bodies.

3 0

3 years ago

Answer the following either true (T) or false (F) (5 pts)

likoan [24]

Answer:

1. True

2. True

3. False

Explanation:

The office location is where the soil layer is not uniform. The thickness of the soil varies which could lead to doors being jammed. The engineer needs to estimate the differential in clay soil.

The inclined surface can hold less weight than a vertical surface. The capacity to hold the weight is due to the gravitational force which is exerted to the load.

6 0

3 years ago

A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such t

bekas [8.4K]

Answer:

a)Δs = 834 mm

b)V=1122 mm/s

$a=450\ mm/s^2$

Explanation:

Given that

$s = 15t^3 - 3t\ mm$

When t= 2 s

$s = 15t^3 - 3t\ mm$

$s = 15\times 2^3 - 3\times 2\ mm$

s= 114 mm

At t= 4 s

$s = 15t^3 - 3t\ mm$

$s = 15\times 4^3- 3\times 4\ mm$

s= 948 mm

So the displacement between 2 s to 4 s

Δs = 948 - 114 mm

Δs = 834 mm

We know that velocity V

$V=\dfrac{ds}{dt}$

$\dfrac{ds}{dt}=45t^2-3$

At t= 5 s

$V=45t^2-3$

$V=45\times 5^2-3$

V=1122 mm/s

We know that acceleration a

$a=\dfrac{d^2s}{dt^2}$

$\dfrac{d^2s}{dt^2}=90t$

a= 90 t

a = 90 x 5

$a=450\ mm/s^2$

4 0

3 years ago

What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin

DochEvi [55]

Answer:

$m_{LP}=0.45\,kg$

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

$Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]$