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3 years ago

13

what is the method of the slope stabilization, and how many type of method of the stabilization, please help explain the types o

f the methods clearly? thnk u

1 answer:

Alik [6]3 years ago

5 0

Answer:

1.Plant Grass and Shrubs. Grass and shrubs are very effective at stopping soil erosion. ...

2.Use Erosion Control Blankets to Add 3.Vegetation to Slopes. ...

4.Build Terraces. ...

5.Create Diversions to Help Drainage

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A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude

Fofino [41]

Answer:

$u = 260.22m/s$

$S_{max} = 1141.07ft$

Explanation:

Given

$S_0 = 89.6ft$ --- Initial altitude

$S_{16.5} = 0ft$ -- Altitude after 16.5 seconds

$a = -g = -32.2ft/s^2$ --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

$S = ut + \frac{1}{2}at^2$

The final altitude after 16.5 seconds is represented as:

$S_{16.5} = S_0 + ut + \frac{1}{2}at^2$

Substitute the following values:

$S_0 = 89.6ft$ $S_{16.5} = 0ft$ $a = -g = -32.2ft/s^2$ and $t = 16.5$

So, we have:

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2$

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45$

$0 = 89.6 + 16.5u- 4383.225$

Collect Like Terms

$16.5u = -89.6 +4383.225$

$16.5u = 4293.625$

Make u the subject

$u = \frac{4293.625}{16.5}$

$u = 260.21969697$

$u = 260.22m/s$

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

$v=u + at$

At the maximum height:

$v =0$ --- The final velocity

$u = 260.22m/s$

$a = -g = -32.2ft/s^2$

So, we have:

$0 = 260.22 - 32.2t$

Collect Like Terms

$32.2t = 260.22$

Make t the subject

$t = \frac{260.22}{ 32.2}$

$t = 8.08s$

The maximum height is then calculated as:

$S_{max} = S_0 + ut + \frac{1}{2}at^2$

This gives:

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2$

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22$

$S_{max} = 89.6 + 260.22 * 8.08 - 1051.11$

$S_{max} = 1141.0676$

$S_{max} = 1141.07ft$

Hence, the maximum height is 1141.07ft

8 0

3 years ago

In python/Spyder

lianna [129]

Answer:

Answer to both the parts A and B are provided in the word document attached.

Explanation:

The explanation of the programs is provided in the attached file along with the screenshot of ATM application output.

8 0

3 years ago

150 lb of force is applied at the end of a shaft that is 2 1/2 ft long. It is applied

Alona [7]

Answer:

Torque: 500

Power: 1700

Explanation:

Since the force is 150 lbs and if the shaft is rotating at 350 rpms, then you will need 1700 for the power in order to keep it running.

5 0

3 years ago

Consider a 8-m-long, 8-m-wide, and 2-m-high aboveground swimming pool that is filled with water to the rim. (a) Determine the hy

Stolb23 [73]

Answer:

The hydrostatic force of 313920 N is acted on each wall of the swimming pool and this force is acted at 1 m from the ground. The hydrostatic force is quadruple if the height of the walls is doubled.

<u>Explanation:</u>

To calculate force on the walls of swimming pool whose dimensions are given as <em>8-m-long, 8-m-wide, and 2-m-high</em>. We know that formula for hydrostatic force is $\text {hydrostatic force}=\text {pressure} \times \text {area,}=\rho g h \times(l \times h)$

$\equiv \rho g h^{2} l$ , we know ρ=density of fluid=1000 $g / c m^{3}$ ,

g=acceleration due to gravity=9.81 $m / s^{2}$ , h=height of the pool=2 m and l=length of the pool=8 m.

hydrostatic force on each wall= $1000 \times 9.81 \times 2^{2} \times 8$ = 313920 N.

<em>The distance at which hydrostatic force is acted is half of the height of the swimming pool. </em>

At 1 m from the ground this hydrostatic force is acted on each wall.

The force is <em>quadruple if the height of the walls of the pool is doubled</em> this is because, the<em> height is doubled and taken as h=4 m</em> and substitute in the equation = $\rho g h^{2} l$ = $1000 \times 9.81 \times 4^{2} \times 8$ = 1255680 N. This is 4 times 313920 N.

5 0

3 years ago

If you are involved in a collision where there is injury, you must report the incident within .......

Misha Larkins [42]

Answer:

24 hours

Explanation:

you must exchange insurance details after a collision if someone is injured. Otherwise you must report the collision to us as soon as possible (and no later than 24 hours). Although you must report such a collision straight away you should always seek medical help in the first instance.

4 0

3 years ago