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3 years ago

Which is a land management technique that limits topsoil loss?

2 answers:

5 0

<span><span>strip-cropping:
cultivation of crops in strips following the contours of the land to minimize erosion</span><span>
</span></span>

ANTONII [103]3 years ago

3 0

the correct answer is strip-cropping

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The theory that light exists in quantized "chunks" was proposed in response

Setler [38]

Its D because of the emission spectra

5 0

3 years ago

Read 2 more answers

Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The ga

Radda [10]

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

6 0

3 years ago

What % of an object’s mass is below the water line if the object’s density is 0.63g/ml?

tiny-mole [99]

<u>Answer:</u>

According to the <u>Archimedes’ Principle</u>, <em>a body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.</em>

In this case we have an object partially immersed in water.

If we observe in <u>the image attached</u>, this object does not completely fall to the bottom because the net force acting on it is zero, this means <u>it is in equilibrium</u>.

This is due to <u>Newton’s first law of motion:</u> if a body is in equilibrium the sum of all the forces acting on it is equal to zero

$F_{total}=0$ (1)

Here, <u>the object experiences two kind of forces</u>:

A force downwards because of its weight:

Where $m$ is the mass of the object and $g$ the acceleration of gravity, which in this case (on earth) is $9.8\frac{m}{s^{2}}$

And a force applied upwards because of the buoyancy:

<h2> $F_{buoyancy}=d_{w}gV_{i}$ (3) </h2>

Where;

$d_{w}$ is the <u>density of Water</u> which is $1\frac{g}{ml}$

And $V_{i}$ is <u>the volume of the immersed part of the object</u>

Well, if the total net force exerted to the object is zero:

$F_{total}= F_{buoyancy}-W=0$ Note $W$ has a negative sign because this force is applied downwards

This means the buoyancy is equal to the weight:

<h2> $F_{buoyancy}=W$ (4) </h2>

$d_{w}gV_{i}=mg$ (5)

Dividing by $g$ in both sides:

$d_{w}V_{i}=m$ (6)

Now, <u>the density</u> of an object or fluid is the relation between the mass and the volume, in the case of the object of this problem is:

Where $V$ is the <u>Volume of the object</u>.

From (7) we can find the mass of the object:

and substitute this value in (6), in order to get a relation between the density of the object and the density of the water, as follows:

$d_{w}V_{i}= dV$

$\frac{V_{i}}{V} =\frac{d}{ d_{w}}$ (9)

Then;

This result is the same as $\frac{63}{100}$ which represents the $63\%$

Therefore the answer is:

<h2> $63\%$ of the mass of the object is below the water line or <u>immersed in water </u></h2>

5 0

3 years ago

What source of renewable energy makes up the vast majority of current global renewable energy resource use, and why is this?

umka2103 [35]

Answer:

hydropower

Explanation:

hydroelectric power is a domestic source of energy. This automatically allows each state to produce their own energy without being reliant on international fuel sources. It also won't pollute the air like power plants that burn fossil fuels, such as coal or natural gas. This is because it is powered by water.

I hope this helped. sorry it took so long. I was trying to get the right resources 4 you

8 0

3 years ago

Horizontal circular motion questions:

NemiM [27]

1. D, constant
Force must be in the direction of the velocity to change speed.

2. B. to the center of the curvature
This allows the vehicle to complete the turn.

7 0

3 years ago