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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 school bus has a mass (including the driver and passengers) of 1.64 times 10^4 kg and is driving north at a speed of 15.2 km/h

Butoxors [25]

Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is $m_1=1.6\times 10^4 kg$

speed of bus towards North $v_1=15.2 km/h\approx 4.22\ m/s$

mass of bus travelling in South direction is $m_2=1.578\times 10^4 kg$

speed of bus $v_2=12.2 km/h\approx 3.38\ m/s$

mass of each Passenger in south moving bus $m_0=64.8 kg$

Momentum of North moving bus

$P_1=m_1\times v_1$

$P_1=1.6\times 10^4\times 4.22$

$P_1=6.768\times 10^4 kg-m/s$

Momentum with south moving bus

$P_2=m_2\times v_2+n\cdot m_0\times v_2$

$P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38$

For total momentum to be towards south

$P_2-P_1$ should be greater than 0

thus for least value of n

$P_2=P_1$

$(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4$

$1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4$

$n=\frac{4243.6686}{64.8}=65.48\approx 66$

5 0

3 years ago

A very small object with mass 8.30×10-9 kg and positive charge 6.90×10-9 C is projected directly toward a very large insulating

Rainbow [258]

Answer:

$41.4496148484\ m/s$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

$\sigma$ = Surface charge density = $5.9\times 10^{-8}\ C/m^2$

$\Delta x$ = 0.57-0.26

q = Charge = $6.9\times 10^{-9}\ C$

m = Mass of object = $8.3\times 10^{-9}\ kg$

Electric field due to a sheet is given by

$E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m$

Electric field is given by

$E=\dfrac{V}{d}$

Voltage is given by

$V=E\Delta x$

Kinetic energy is given by

$K=qV$

$\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s$

The initial speed of the object is $41.4496148484\ m/s$

7 0

3 years ago

he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug

S_A_V [24]

Answer:

207.4 N

Explanation:

The torque $\tau$ on a body is

$\tau = r* F$ where r is the radius vector from the point of rotation to the point at which force F is applied.

The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.

Therefore, torque is also given by

$\tau = rF\sin \theta$

Where $\theta$ is the angle between r and F.

Use the expression of torque.

Substitute L for r in the equation $\tau = rF\sin \theta$

$\tau = LF\sin \theta$

Where L is the length of the wrench.

Making F the subject

$F = \frac{\tau }{{L\sin \theta }}$

Force required to pull the wrench is given as,

$F = \frac{\tau }{{L\sin \theta }}$

Substitute $47{\rm{ N}} \cdot {\rm{m}}$ for $\tau$ , 25 cm for L, and 115o for $\theta$

$\begin{array}{c}\\F = \frac{{47{\rm{ N}} \cdot {\rm{m}}}}{{\left( {25{\rm{ cm}}} \right)\sin {{115}^{\rm{o}}}}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 207.435{\rm{ N}}\\\\ \approx 207.4{\rm{ N}}\\\end{array}$

6 0

3 years ago

100 °C is a greater temperature than which of the following?

ohaa [14]

When somebody hands you a Celsius°, it's easy to find the equivalent Fahrenheit°.

Fahrenheit° = (1.8 · Celsius°) + 32° .

So 100°C works out to 212°F.

It's also easy to find the equivalent Kelvin. Just add 273.15 to the Celsius.

So now you can see that 100°C is equal to A and D,
and it's less than B .

The only one it's greater than is C .

6 0

3 years ago

what volume of alcohol will have the same mass as 4.2m^3 of petrol? (density of alcohol 0.4kg/m^3 and petrol is 0.3kg/m^3)

Alexxx [7]

Answer:

3.15m³

Explanation:

To solve this problem, let us first find the mass of the petrol from the given dimension.

Mass = density x volume

Volume of petrol = 4.2m³

Density of petrol = 0.3kgm⁻³

Mass of petrol = 4.2 x 0.3 = 1.26kg

So;

We can now find the volume of the alcohol

Volume of alcohol = $\frac{mass}{density}$

Mass of alcohol = 1.26kg

Density of alcohol = 0.4kgm⁻³

Volume of alcohol = $\frac{1.26}{0.4}$ = 3.15m³

7 0

3 years ago