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4 years ago

Which of the following changes to the local hydrologic cycle will result if large areas of land experienced clear-cutting?

Physics

2 answers:

mariarad [96]4 years ago

8 0

Answer:

Evaporation of water would increase

Explanation:

This is because clear cutting would increase pore space in the soil which would increase ease of evaporation of water, and removal of plant canopy over earth thereby decreases shading of soil and increasing evaporation.

lutik1710 [3]4 years ago

7 0

Answer:

The correct answer is "Evaporation of water from the soil will increase".

Explanation:

Evaporation is the step of the hydrologic cycle at which deposits of liquid water got into the atmosphere by transforming into a water vapor. The level of evaporation is closely related with the surface area of foliage, where the more surface area of foliage the less evaporation will occur. Therefore, if large areas of land experience clear-cutting, evaporation of water from the soil will increase.

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What is the mass of a liquid having a density of 1.50 g/ml and a volume of 3.5 liters?

PIT_PIT [208]

We know the formula for density = Mass/ volume

So Mass, M = Volume * Density

Volume = 3.5 L= 0.0035 $m^3$

Density = 1.50 g/ml = 1500 $kg/m^3$

Mass, M = 0.0035*1500 = 5.25 kg

So mass of liquid having density 1.50 g/ml and volume 3.5 liters is 5.25 kg.

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4 years ago

The mass number of Fe2+ is 56. How many neutrons are there in a single Fe2+ atom?

bonufazy [111]

<span>The atomic number of a neutral atom is equal to the number of protons and the number of electrons of the atom. The atomic weight meanwhile is equal to the sum of the number of protons and number of neutrons. In this case, the number of protons is equal to 26. The number of neutrons hence is 30. </span>

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3 years ago

Read 2 more answers

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$