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3 years ago

Which of the following would most likely be considered nonpoint source pollution

1 answer:

8 0

The best answer is b) increased turbidity from erosion.

Nonpoint source pollution generally happens as a result of many systems interacting, and is not directly attributed to one event or pollutant. Generally, natural environmental systems participate in pollution of this kind, regardless of whether or not human activity was a factor. Examples include water runoff, or erosion.

The other pollutants listed have a direct cause and direct effect, the animal waste goes directly from the animals to the ground they live on, the car shop directly sumps the oil on the ground, and the oil tank leaks directly into the earth. Erosion causing turbidity is a less direct form of pollution, and is due to the synthesis of several natural phenomena

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A bob attached to a string of length L = 1.25 m, initially found at the equilibrium

malfutka [58]

The maximum displacement angle of the bob is 13⁰.

The given parameters;

Length of the pendulum, L = 1.25 m
Initial velocity of the bob, v = 0.8 m/s

The maximum displacement of the bob is calculated by applying the principle of conservation of energy;

$P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m$

The maximum displacement angle is calculated as follows;

$cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0$

Thus, the maximum displacement angle of the bob is 13⁰.

Learn more here:brainly.com/question/13981780

4 0

2 years ago

Is O2 considered one atom or 2 atoms?

natulia [17]

O2 is considered 2 atoms because O2 is 2 oxygen atoms.

4 0

3 years ago

Read 2 more answers

Nine-year-old Dakoda realizes that the quantity of water in a glass remains the same, even when the water is poured into a diffe

Archy [21]

Answer:

Conservation

Explanation:

She has observation conservation because If the temperature of the liquids stays constant and the container is insulated and not heat or cool the liquid much would not change the density of the liquid very much so that it's original volume could remain constant.

The interesting thing is not that the child assumes the taller glass holds more liquid but that they fail to understand conservation: the fact that the water from one glass is going to be the same amount after being poured into any other container. It's as if they did not realize the water came from the same glass.

3 0

3 years ago

A,b,c,d please I don't understand it.

emmasim [6.3K]

For A: LED outdoor lighting.
Because you want it to be able to withstand outdoor conditions. If it can’t withstand... there’s no light... no light... no advertisement.

For B: it would be lithonium lights. It’s found in most offices and gives the best light for working day or night.

For C: you’ll want elevated lights so they are noticed for the plane to land and for people and others to see.

For D: you want white light and you want it to also be able to withstand outdoor conditions. Cars and people need to see at night, especially at night. Solar would be a great choice.

Hope this helps! I tried hard.

7 0

2 years ago

A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire

Morgarella [4.7K]

Answer:

electric field in the wide wire is

E₂ = $\frac{E}{4}$

Explanation:

given

length of the copper wire = L

radius of the copper wire r₁ = b

length of the second copper wire = L

radius of the second copper wire r₂ = 2b

electric field in the narrow wire = E₁=E

recall

resistance R = ρL/A

where ρ is resistivity of the copper wire, L is the length, and A is the cross sectional area.

Resistance of narrow wire, R₁

R₁ = ρL/A

where A = πb²

R₁ = ρL/πb²---------- eqn 1

Resistance of wide wire, R₂

R₂ = ρL/A

where A = π(2b)²

R₂ = ρL/π(2b)²

R₂ = ρL/4πb²-------------- eqn 2

R₂ = ¹/₄(ρL/πb²)

comparing eqn 1 and 2

R₁ = 4R₂

calculating the current in the wire,

I = E/(R₁ + R₂)

recall

R₁ = 4R₂

∴ I = E/(4R₂ + R₂)

I = E/5R₂

calculating the potential difference across R₁ & R₂

V₁ = IR₁

I = E/5R₂

∴ V₁ = ER₁/5R₂

R₁ = 4R₂

V₁ = 4ER₂/5R₂

∴V₁ = ⁴/₅E

potential difference for R₂

V₂= IR₂

I = E/5R₂

∴ V₂ = ER₂/5R₂

V₂ = ER₂/5R₂

∴V₂ = ¹/₅E

so, electric field E = V/L

for narrow wire E₁ = V₁/L ----------- eqn 3

for wide wire, E₂ = V₂/L------------ eqn 4

compare eqn 3 and 4

E₂/E₁ = V₂/V₁( L is constant)

E₂/E₁ = ¹/₅E/⁴/₅E

E₂ = E₁/4

note E₁ = E

∴E₂ = $\frac{E}{4}$

8 0

3 years ago