All categories

3 years ago

What determines the depth of the foundation excavation

Engineering

2 answers:

WARRIOR [948]3 years ago

7 0

Answer:

The required depth of any foundation can depend on several factors: Soil bearing capacity. This determines how much load (weight or force) the existing soil can withstand. Soil type.

Explanation:

Klio2033 [76]3 years ago

6 0

They determine based on ground movement studies in that area

You might be interested in

Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area

dybincka [34]

Answer:153.76 MPa

Explanation:

$Initial Area\left ( A_0\right )=12.56 mm^2$

$Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2$

$Die angle=30^{\circ}$

$\alpha =\frac{30}{2}=15^{\circ}$

$\mu =0.08$

$Yield stress\left ( \sigma _y \right )=350 MPa$

$B=\mu cot\left ( \aplha\right )=0.2985$

$\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B$

$\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}$

$\sigma _{pressure}=153.76 MPa$

8 0

3 years ago

Which sentence is correct about the exergy of an empty (pressure around zero Pascal) tank with a volume of V, located in an envi

sineoko [7]

Answer:

The correct option is;

c. the exergy of the tank can be anything between zero to P₀·V

Explanation:

The given parameters are;

The volume of the tank = V

The pressure in the tank = 0 Pascal

The pressure of the surrounding = P₀

The temperature of the surrounding = T₀

Exergy is a measure of the amount of a given energy which a system posses that is extractable to provide useful work. It is possible work that brings about equilibrium. It is the potential the system has to bring about change

The exergy balance equation is given as follows;

$X_2 - X_1 = \int\limits^2_1 {} \, \delta Q \left (1 - \dfrac{T_0}{T} \right ) - [W - P_0 \cdot (V_2 - V_1)]- X_{destroyed}$

Where;

X₂ - X₁ is the difference between the two exergies

Therefore, the exergy of the system with regards to the environment is the work received from the environment which at is equal to done on the system by the surrounding which by equilibrium for an empty tank with 0 pressure is equal to the product of the pressure of the surrounding and the volume of the empty tank or P₀ × V less the work, exergy destroyed, while taking into consideration the change in heat of the system

Therefore, the exergy of the tank can be anything between zero to P₀·V.

6 0

3 years ago

Engineers create a new metal that is stronger than steel but much lighter. This material is also significantly cheaper than what

zysi [14]

The best step for the engineers to make next is option D. Begin to design an airplane using this metal.

<h3>What is the metallic is plane parts?</h3>

Aluminum and its alloys are nevertheless very famous uncooked substances for the production of business planes, because of their excessive electricity at exceedingly low density. Currently, excessive-electricity alloy 7075, which includes copper, magnesium and zinc, is the only used predominantly withinside the plane industry.

The solution is D, due to the fact even as it's far crucial to marketplace the fabric and ensure humans are inquisitive about buying, they first want to attempt to layout aircraft the usage of this fabric. There isn't anyt any use promoting an aircraft constituted of this material_ if a aircraft can not be built.

Read more about the aircraft:

brainly.com/question/5055463

#SPJ1

7 0

2 years ago

The total floor area of a building, including below-grade space but excluding unenclosed areas, measured from the exterior of th

alex41 [277]

Answer:

Gross building area

Explanation:

The Gross building area refers to the entire area of a building covering all the floors. The measurement is expressed in square feet. The Gross building area also includes basements, penthouses, and mezzanines. It is calculated by estimating the exterior dimension of the building. Storage rooms, laundries, staircases are also a part of the gross building area.

6 0

3 years ago

While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr),

MAXImum [283]

Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

$y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Where:

$y_{o}$ , $y$ - Initial and final vertical position, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$\theta$ - Launch angle, measured in sexagesimal degrees.

$g$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $y_{o} = 2\,m$ , $y = 0\,m$ , $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $g = -9.807\,\frac{m}{s^{2}}$ , then the time taken by the ball is:

$-4.904\cdot t^{2}+13.482\cdot t +2 = 0$ (2)

This second order polynomial can be solved by Quadratic Formula:

$t_{1} \approx 2.890\,s$ and $t_{2} \approx -0.141\,s$

Only the first root offers a solution that is physically reasonable. That is, $t \approx 2.890\,s$ .

The vertical velocity of the ball is calculated by this expression:

$v_{y} = v_{o}\cdot \sin \theta +g\cdot t$ (3)

Where:

$v_{o,y}$ , $v_{y}$ - Initial and final vertical velocity, measured in meters per second.

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t \approx 2.890\,s$ , then the final vertical velocity is:

$v_{y} = -14.860\,\frac{m}{s}$

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball ( $x$ ) is determined by the following expression:

$x = (v_{o}\cdot \cos \theta)\cdot t$ (4)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $t \approx 2.890\,s$ , then the distance covered by the ball is:

$x = 32.695\,m$

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground ( $v$ ), measured in meters per second, is determined by the following Pythagorean identity:

$v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}}$ (5)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , then the magnitude of the velocity of the ball is:

$v \approx 18.676\,\frac{m}{s}$ .

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

$\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}$

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta_{o} = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , the angle of the total velocity of the ball just before hitting the ground is:

$\theta \approx -52.717^{\circ}$

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

3 0

3 years ago

Read 2 more answers