8. Find the volume of the figure shown below: * V=L x W x H 7 cm 2 cm 2 cm

When a retaining structure moves towards the soil backfill, the stress condition is called:__________.

Alecsey [184]

Answer:

(C) passive state.

Explanation:

The earth pressure is the pressure exerted by the soil on the shoring system. They are three types of earth pressure which are:

a) Rest state: In this state, the retaining wall is stationary, this makes the lateral stress to be zero.

b) Active state: In this state, the wall moves away from the back fill, this leads to an internal resistance. Hence the active earth pressure is less than earth pressure at rest

c) Passive state: In this state the wall is pushed towards the back fill, this leads to shearing resistance. Hence, the passive earth pressure is greater than earth pressure at rest

6 0

3 years ago

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

viktelen [127]

Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

Let's solve for $\theta$

$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$

$\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}$

Or equivalently:

$\theta_2=90^o-\theta_1$

Given Vo=37 m/s and R=70 m

$\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}$

$\theta_1=15^o$

And

$\theta_2=90^o-15^o=75^o$

5 0

3 years ago

Create a variable pounds to store weight in pounds. Convert this to kilograms and assign the result to a variable kilos. The con

vodka [1.7K]

Answer:

>>pounds=13.2

>>kilos=pounds/2.2

Explanation:

Using Matlab to write the program, consider at any time when the weight in pounds is 13.2 lb, this variable of weight is created in MATLAB by typing >>pounds=13.2. To convert it from lb to Kg, we simply divide it by 2.2 hence the second command to created is kilos. For this, the output of the program will be 6 Kg.

5 0

3 years ago

A simple formula to estimate the upward velocity of a rocket (neglecting the aerodynamic drag) is:

Bingel [31]

Answer:

Test code:

>>u=10;

>>g=9.8;

>>q=100;

>>m0=100;

>>vstar=10;

>>tstar=fzero_rocket_example(u, g, q, m0, vstar)

Explanation:

See attached image

5 0

3 years ago

1) Name 5 factors that can lead to pressure drop in the displacement of a fluid being pumped through a pipeline.

Dominik [7]

Answer:

im so sorry I rlly need these points

Explanation:

6 0

3 years ago

8. Find the volume of the figure shown below: * V=L x W x H 7 cm 2 cm 2 cm​

8. Find the volume of the figure shown below: * V=L x W x H 7 cm 2 cm 2 cm