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3 years ago

Explain at least three ways trees assist in keeping high water quality?

Physics

2 answers:

Andrew [12]3 years ago

8 0

Answer:

One of the parts of the tree is its roots, this serves as the anchorage of the tree to the ground, this also lessen the erosion of both the soils and other solid components of the ground (e.g. rocks, etc). The presence of soil, rocks, etc will hold the water in place longer.

Explanation:

swat323 years ago

3 0

<span>Tree's are vital to the eco-system, they clean the air of co2 which is important for the air and water. Tree's also stop erosion and that is an important step in keeping our water clean. Tree's filter water up through the soil with their extensive root system, making for clean water.</span>

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What is the force when two charged spheres distance is in half

slamgirl [31]

Answer:

When two spheres, each with charge Q, are positioned a distance Rapart, they are attracted to ... doubled, the electric-force between the two spheres

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3 years ago

Describe the relationship of attractive forces between molecules

Ede4ka [16]

<span>Describe the relationship of attractive forces between molecules and the ability of a solvent to dissolve a substance. Solvents can dissolve a substance only if the attraction of the solvent molecules is greater than the attraction between the molecules of the substance.</span>

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3 years ago

Why we don’t feel atmospheric pressure?

Veseljchak [2.6K]

Answer:

The reason we can't feel it is that the air within our bodies (in our lungs and stomachs, for example) is exerting the same pressure outwards, so there's no pressure difference and no need for us to exert any effort.

8 0

3 years ago

Read 2 more answers

A solid ball of radius rb has a uniform charge density rho.

Oksana_A [137]

Note: question B is incomplete.

Complete Question

A solid ball of radius rb has a uniform charge density ρ.

a. What is the magnitude of the electric field E(r) at a distance r>rb from the center of the ball? Express your answer in terms of ρ, rb, r, and ϵ0.

b. What is the magnitude of the electric field E(r) at a distance r<rb from the center of the ball? Express your answer in terms of ρ, r, rb, and ϵ0.

c. Let E(r) represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true?

1. E(0) = 0.

2. E(rb) = 0

3. lim E(r) = 0.

4. The maximum electric field occurs when r = 0.

5. The maximum electric field occurs when r = rb.

6. The maximum electric field occurs as r to infinity.

Answer:

a) the magnitude of E(r)= ρr³/3 ε₀r²

b) the magnitude at distance r from the centre E(r)= ρr/3 ε ₀ ( if r<rb)

c) statements 1(E(0) = 0), 3(E(0) = 0) and 5(The maximum electric field occurs when r = rb.) are true

Explanation:

given

charge density = ρ , ε

Volume of sphere , V = (⁴/₃)πr³

a) charge density = charge/volume

ρ = q ÷ V

make charge the subject of the formula

∴q = ρ × V= ρ× (⁴/₃)πr³

where r³ = rb³(at distance rb³)

recall

E= q/4πε₀r²

E= ρ × (⁴/₃)πrb³/4πε₀r²

∴E(r)= ρrb³/3 ε ₀r²

(b) The Gaussian surface is inside the ball, therefore, surface only encloses a portion of ball's charge .

The net charge enclosed by the Gaussian surface is different to the of net charge enclosed in (a)

Recall

E= q/4πε₀r²

V= (⁴/₃)πr³

E= ρ × (⁴/₃)πr³/4πε₀r²

∴E(r)= ρr/3 ε₀

limr-----∝

E(r)= 0

The maximum electric field occurs when r=rb.

4 0

3 years ago

Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts

podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height =?

$y_{0}$ : is the initial height = 1 m

$v_{0_{y}$ : is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time

First, we need to find the time by using the following equation:

$t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s$

Now, the height is:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m$

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

$0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}$

$1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0$

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

$x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m$

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!

4 0

3 years ago