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3 years ago

What is the most effective way to protect dry lands from desertification

Physics

2 answers:

liberstina [14]3 years ago

5 0

Answer:

The correct answer is Planting a tree belt (C)

Explanation:

I just got it right on my test:)

Sonja [21]3 years ago

4 0

Answer: Planting a tree belt.

Explanation:

The most effective method by which the desertification of the soil can be controlled is by planting more and more trees.

The process of desertification occurs when there is a loss of vegetation, water bodies and animals from a particular area. Loss of trees makes the area look like desert.

So, planting trees along with some irrigation will help to cope with problem.

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A quantity must be divided by multiples of ten when converting from a larger unit to a smaller unit.

sineoko [7]

Answer:

what is the answers? i cant help you without the answers

Explanation:

5 0

3 years ago

The potential-energy function u(x) is zero in the interval 0≤x≤l and has the constant value u0 everywhere outside this interval.

VMariaS [17]

Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):
K = ½mv²
v = √(2K/m)
λ = h/(mv)
= h/(m√(2K/m))
= h/√(2Km)

So λ is proportional to 1/√K.
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:
K = 6U₀

Outer the potential well the potential energy is U₀, so
K = 5U₀
(because kinetic and potential energies add up to 6U₀)

Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is:
1/√(5U₀) : 1/√(6U₀)
= √6 : √5

5 0

4 years ago

The current theory of the structure of the

IRISSAK [1]

1) The mass of the continent is $3.3\cdot 10^{21} kg$

2) The kinetic energy of the continent is 624 J

3) The speed of the jogger must be 4 m/s

Explanation:

We start by finding the volume of the continent. We have:

$L = 5850 km = 5.85\cdot 10^6 m$ is the side

$t = 35 km = 3.5\cdot 10^4 m$ is the depth

So the volume is

$V=L^2 t = (5.85\cdot 10^6)^2 (3.5\cdot 10^4)=1.20\cdot 10^{18} m^3$

We also know that its density is

$d=2750 kg/m^3$

Therefore, we can find the mass by multiplying volume by density:

$m=dV=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg$

The kinetic energy of the continent is given by:

$K=\frac{1}{2}mv^2$

where

$m=3.3\cdot 10^{21} kg$ is its mass

v = 3.2 cm/year is the speed

We have to convert the speed into m/s. We have:

3.2 cm = 0.032 m

$1 year = 1(365)(24)(60)(60)=3.15\cdot 10^7 s$

So, the speed is:

$v=\frac{0.032 m}{3.15 \cdot 10^7 s}=1.02\cdot 10^{-9} m/s$

So, we can now find the kinetic energy:

$K=\frac{1}{2}(1.20\cdot 10^{21})(1.02\cdot 10^{-9})^2=624 J$

Here we have a jogger of mass

m = 78 kg

And the jogger has the same kinetic energy of the continent, so

K = 624 J

The kinetic energy of the jogger is given by

$K=\frac{1}{2}mv^2$

where v is the speed of the jogger.

Solving for v, we find the speed that the jogger must have:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(624)}{78}}=4 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0

3 years ago

(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of

NISA [10]

Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

$F = \dfrac{kq_1q_2}{r^2}$

where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.

8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

$F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}$

Force due to Q₃ :

$F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}$

8.3. The net force on the particle at Q₂ is the vector

$\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}$

Its magnitude is

$\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}$

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

$\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}$

where we subtract 180° because $\vec F$ terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.

8 0

2 years ago

A material that allows an electric current to flow through it is an electrical conductor. A material through which an electric c

ElenaW [278]

An insulator does not have electrons that are free to move, and a conductor does.

4 0

3 years ago