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2 years ago

Which element accumulates in the environment due to the use of

Physics

1 answer:

marissa [1.9K]2 years ago

6 0

To Find :

Which element accumulates in the environment due to the use of fertilizers.

Solution :

Most of the fertilizers used in daily practice are nitrogen fertilizers .

When soil became waterlogged, soil organisms take the oxygen they need from nitrates, leaving the nitrogen in a gaseous form which escapes into the air.

Also, nitrogen is very stable and did not participate in most of the reactions.

So, nitrogen is accumulates in the environment .

Therefore, option C. is correct.

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Your cousin Jannik skis down a blue square ski slope, with an initial speed of 3.6 m/s. He travels 15 m down the mountain side b

fenix001 [56]

Answer: The loss of energy due to friction is equal to 1,253 J.

Explanation:

The problem tells us that the skier has an initial speed of 3.6 m/s, which means that his initial kinetic energy is as follows:

K₁ = 1/2 m v₁² = 1/2 . 58.0 Kg. (3.6)² (m/s)² = 376 J

After coming to a flat landing, his final speed is 7.8 m/s, so the final kinetic energy is as follows:

K₂ = 1/2 m v₂² = 1/2. 58.0 Kg. (7.8)² (m/s)² = 1,764 J

Now, when skying down the slope the increase in kinetic energy only can come from another type of energy, in this case, gravitational potential energy.

If we take the ground flat level as a Zero reference, the initial gravitational potential energy, can be written as follows, by definition:

U₁ = m.g. h (1)

Now, we don't know the value of the height h, but we know that the incline has a 18º angle above the horizontal, and that the distance travelled along the incline is 15 m.

By definition, the sinus of an angle, is equal to the proportion between the height and the hypotenuse , so we can write the following equation:

sin 18º = h / 15 m ⇒ h = 15 m. sin 18º = 4.6 m

Replacing in (1), we get:

U₁ = 58.0 Kg. 9.8 m/s². 4.6 m = 2,641 J

So, we can get the total initial mechanical energy, as follows:

E₁ = K₁ + U₁ = 376 J + 2,641 J = 3,017 J

After arriving to the flat zone, all potential energy has become in kinetic energy, even though not completely, due to the effect of friction.

This remaining kinetic energy can be written as follows:

E₂ = K₂ = 1,764 J

The difference E₂-E₁, is the loss of energy due to friction forces acting during the travel along the 15 m path, and is as follows:

ΔE= E₂ - E₁ = 1,764 J - 3,017 J = -1,253 J

8 0

2 years ago

10. Discrimination refers to behaviors that

salantis [7]

Answer:

Explanation:

the only one that makes sense

4 0

3 years ago

During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±

Alla [95]

Answer:

P = 1 (14,045 ± 0.03 ) k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

Δ (Pₓ / Py) =?

Let's start by finding the momentum of each vehicle

car X

Pₓ = m vₓ

Pₓ = 2.34 2.5

Pₓ = 5.85 kg m

car Y

Py = 2,561 3.2

Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

ΔPₓ = m Δv + v Δm

ΔPₓ = 2.34 0.01 + 2.561 0.01

ΔPₓ = 0.05 kg m

Δ $P_{y}$ = m Δv + v Δm

ΔP_{y} = 2,561 0.01+ 3.2 0.001

ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

P = Pₓ / $P_{y}$

ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

ΔP = 0.006 + 0.0026

ΔP = 0.009 kg m

The result is

P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s

7 0

3 years ago

In our solar system, the most likely planet (other than Earth) to have life on it is currently thought to be

VLD [36.1K]

Answer: Mars

Explanation:

Mars is a planet similar to the earth which the rest can't sustain life either because it's too hot or cold or too much gas maybe even toxic acid.

6 0

3 years ago

You stand on a frictionless platform that is rotating with an angular speed of 5.1 rev/s. Your arms are outstretched, and you ho

timurjin [86]

Answer:

$\omega'=19.419\ rev.s^{-1}$

Explanation:

Given:

angular speed of rotation of friction-less platform, $\omega=5.1\ rev.s^{-1}$

moment of inertia with extended weight, $I=9.9\ kg.m^2$

moment of inertia with contracted weight, $I'=2.6\ kg.m^2$

<u>Now we use the law of conservation of angular momentum:</u>

$I.\omega=I'.\omega'$

$9.9\times 5.1=2.6\times \omega'$

$\omega'=19.419\ rev.s^{-1}$

The angular speed becomes faster as the mass is contracted radially near to the axis of rotation.

5 0

3 years ago