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### Journal ofEcology and The Natural Environment

• Abbreviation: J. Ecol. Nat. Environ.
• Language: English
• ISSN: 2006-9847
• DOI: 10.5897/JENE
• Start Year: 2009
• Published Articles: 394

## Deforestation impact on ecosystem services in Kamfinsa sub-catchment of Kafue River Basin in Zambia

##### Deuteronomy Kasaro
• Deuteronomy Kasaro
• Department of Soil Science, School of Agricultural Sciences, University of Zambia, Lusaka, Zambia.
##### Elijah Phiri
• Elijah Phiri
• IWRM Center, School of Mines, University of Zambia, Lusaka, Zambia.
##### Imasiku Nyambe
• Imasiku Nyambe
• Department of Soil Science, School of Agricultural Sciences, University of Zambia, Lusaka, Zambia.
Forests are important in the lives of local communities as they provide poles for construction, wood fuel for cooking, help reduce soil erosion and provide various foods and medicines due to biodiversity. Most local communities depend on forests in supplementing their livelihoods. At the moment, forests are being lost and it is important to understand the impacts of the lost. Therefore, the objective of the study was to provide a holistic evaluation of the impacts of deforestation on ecosystem services in Kamfinsa sub-catchment of Kitwe in Zambia. Loss of forest biomass, biodiversity, and soil erosion were key proxy variables for loss of ecosystem services. Geographical Information System (GIS) and Remote sensing techniques were used to assess the impacts on the ecosystem. The results showed that deforestation reached 576.3 ha per year during the period studied. The forest loss corresponds to an emission of 43.73 ton of carbon per hectare from above and below ground biomass valued at US$243.60 per hectare. According to this research soil erosion risk assessment, 1.59 ton of soil was lost per hectare per year, equivalent to US$ 57.20 loss per hectare. The indigenous forest cover reduced from 13,430.5 ha (1990) to 2,904.7 ha (2010), with a corresponding change in NDVI index for loss of forest vigor and biodiversity of 0.56 and 0.32, respectively. The major forest loss occurred from indigenous forests. The study has shown that deforestation in Kamfinsa sub-catchment area calls for the urgent promotion of an integrated and comprehensive approach to addressing the drivers of deforestation to ensure continued supply of ecosystem services.


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## electric field strength and surface charge densitys

The figure shows two very large slabs of metalthat are parallel and distance L. Each slab has a total surfacearea (top + bottom) .The thickness of each slab is so small in comparison to its lateraldimensions that the surface area around the sides is negligible.Metal has total charge and metal has total charge . Assume is positive.
Determine the electric field strengths in all 5regions.
Also, find the surface charge densitys in all 4surfaces

•
When the two plates are charged with the charge of thesign, all charge on either plate will move to the outer surfaces ofeach plate because of repulsion from the charge on the otherplate. All charge Q on metal 1 will now move to its topsurface 'a' and all charge 2Q on metal 2 will move  to itsbottom surface d.
Now apply Gauss law by making Gaussian pill box cylinders asfollows. In the picture you should imagine the pill boxes to beidentical thin cylinders with end surface areas dA

Remember with the above configuration, all electricfield inside of either plate is zero.
Applying Gauss law to pill box on surface a of plate 1, weget
E*dA + 0*dA  =
which gives E =
For the bottom area of plate 2, the electric fieldwill be twice the above due to the 2Q total chargeit had to start with.
What about region 3 between the plates? Apply Gauss law topill boxes for either plate should give you zero electric fieldthere due to no charge being enclosed by pill boxes.
Clearly the surface charge density on area a is  charge Qdivided by A/2. (A is total area top +bottom)
Surface charge density on area d =  2Q divided byA/2
Get homework help

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# getData

Get current values of tunable-surface coefficients

## Syntax

``Kco = getData(K)``
``KcoJ = getData(K,J)``

## Description

````Kco = getData(K)` extracts the current value of a tunable surface. `K` is a `tunableSurface` object that represents the parametric gain surface:$K\left(n\left(\sigma \right)\right)=\gamma \left[{K}_{0}+{K}_{1}{F}_{1}\left(n\left(\sigma \right)\right)+\dots +{K}_{M}{F}_{M}\left(n\left(\sigma \right)\right)\right],$where: σ is a vector of scheduling variables.n(σ) is a normalization function (see the `Normalization` property of `tunableSurface`). γ is a scaling factor (see the `Normalization` property of `tunableSurface`).F1,...,FM are basis functions.K0,...,KM are tunable coefficients.`getData` evaluates `K` at the current values of the coefficients K0,...,KM.```
````KcoJ = getData(K,J)` extracts the current value of the coefficient of the Jth basis function FJ. Use `J` = 0 to get the constant coefficient K0.```

## Input Arguments

collapse all

Gain surface, specified as a `tunableSurface` object,

Index of basis function, specified as a nonnegative integer. To extract the constant coefficient K0, use `J` = 0.

## Output Arguments

collapse all

Current coefficient values of the tunable surface, returned as an array.

If the tunable surface `K` is a scalar-valued gain, then the length of `K` is `(M+1)`, where `M` is the number of basis functions in the parameterization. For example, if `K` represents the tunable gain surface:

`$K\left(\alpha ,V\right)={K}_{0}+{K}_{1}\alpha +{K}_{2}V+{K}_{3}\alpha V,$`

then `Kco` is the 1-by-4 vector [K0,K1,K2,K3].

For array-valued gains, each coefficient expands to the I/O dimensions of the gain. These expanded coefficients are concatenated horizontally in `Kco`. (See `tunableSurface`.) For example, for a two-input, two-output gain surface, `Kco` has dimensions `[2,2(M+1)]`.

Coefficient of the Jth basis function in the tunable surface parameterization, returned as a scalar or an array.

If the tunable surface `K` is a scalar-valued gain, then `KcoJ` is a scalar. If `K` is an array-valued gain, then `KcoJ` is an array that matches the I/O dimensions of the gain.


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Warren Velez

2023-02-27

When the equation y=4x−10 is written in the standard form, the value of ${c}^{2}$ will be.

objesilo5lr1

Given equation: y=4x−10
Also, y=4x−10 can be written as 4x−y−10=0
So, on compairing 4x−y−10=0 to the standard form ax+by+c=0 we get,
a=4
b=−1
c=−10
Now,
${c}^{2}=\left(-10{\right)}^{2}$
${c}^{2}=100$
Hence, value of ${c}^{2}$ is 100.

Do you have a similar question?


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# Is there Even any Area?

Geometry Level pending

I have a unit circle, which has a central angle of $$27.5^{\circ}$$. There are two points formed by the central angle B and B', which have lines tangent to both points. What is the area between the circle and the two tangent lines? (Round the answer to four decimal places.)

×


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# Probability question in telephone network failures

Hi,

I have this problem

## Homework Statement

In telephone network, an average number of failures during a month for one telephone owner is 8. What is the probability that some telephone owner will experience more than 4 failures?

## The Attempt at a Solution

I have no idea what distribution should I use to model this problem. So far I've come up with binomial distribution with probability of individual event 8/k, where k is number of connection the telephonist will make during a month.

But that seems really odd to me...

Thank you.

AlephZero
Homework Helper
So you have the situation where there are k trials (attempted calls), the probability of "success" (i.e. there is a fault) is 8/k, and k is probably large.

What distribution is an approximation to the Binomial(n,p) as n becomes large and np stays constant?

So you have the situation where there are k trials (attempted calls), the probability of "success" (i.e. there is a fault) is 8/k, and k is probably large.

What distribution is an approximation to the Binomial(n,p) as n becomes large and np stays constant?

One way is to approximate it with Poisson distribution, but can I solve it using Central Limit Theorem? Which one is better?

I tried to finish it using the Poisson distribution and it seems ok.

Anyway, using Central Limit Theorem I got result depending on n, is it ok?

Yeap, I think Poisson distribution is more appropriate here.