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mario62 [17]
3 years ago
5

How much energy from the sun actually reaches the corn answer?

Physics
2 answers:
professor190 [17]3 years ago
7 0

Answer:

It depends upon the sunlight intensity known as ALBEDO, the amount of energy used by photosynthesis and the chemical composition of corn.

Explanation:

Albedo is a measure of the amount of light reflected from an object. Albedo is normally expressed as a decimal value representing the percentage of light reflected.

Photosynthesis, a biological process, uses the energy of sunlight to manufacture sugar, which serves as the universal food for life. Oxygen produced as a product of photosynthesis is released into the environment.

The formula for this reaction is:

6CO2 + 6H2O + sunlight energy → C6H12O6 + 6O2

• where CO2 represents carbon dioxide

• H2O represents water

• C6H12O6 represents the sugar molecules (carbohydrate)

• and O2 represents oxygen


Serhud [2]3 years ago
3 0
The energy from the sun that reaches the corn is about two billionths.
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Tap on the photo. For each diagram, explain why the light behaves in the way that it does.
dem82 [27]

Answer:

Diagram 1, 3 and 4 can be explained with the phenomenon of refraction.

Refraction occurs when a ray of light crosses the interface between two mediums with different optical density: when this occurs, the ray of light is bent and its speed changes, according to Snell's law

n_1 sin \theta_1 = n_2 sin \theta_2

where n_1,n_2 are the refractive index of the 1st and 2nd medium

\theta_1, \theta_2 are the angle that the incident ray and the refracted ray makes with the normal to the interface

In diagram, 1, the ray of light arrives perpendicularly to the interface, so it is refracted through the medium but it doesn't change its direction (only its speed).

In diagram 3, the ray of light is refracted twice: at the 1st interface and at the 2nd interface. In the 1st case, it goes from a medium with lower refractive index to a medium with higher refractive index (n_1), this means that \theta_2, so the ray bends towards the normal. Vice-versa, in the 2nd case the ray goes from a medium with higher refractive index to a medium with lower refractive index (n_1>n_2), so it bends away from the normal (\theta_2>\theta_1).

In diagram 4, the ray of light is also refracted twice. The ray of light here acts exactly the same as in diagram 3, h

However, this time the 2nd interface is the opposite direction with respect to diagram 3, so in this case the ray of light at the 2nd interface bends in the opposite direction (still away from the normal).

Diagram 2 instead is an example of reflection, that occurs when a ray of light bounces off the interface between the two mediums, withouth entering the 2nd medium.

According to the law of reflection:

- The incoming ray, the reflected ray and the normal to the boundary are all in the same plane

- The angle of incidence is equal to the angle of reflection (both are measured relative to the normal to the boundary)

Therefore in this diagram, the ray of light hits the boundary at approx. 45 degrees from the normal, and then it is reflected back approximately at 45 degrees on the other side with respect to the normal.

3 0
4 years ago
Qs # 1. A satellite orbits earth with a mean altitude of 361 Km. If the orbit is
alisha [4.7K]

Answer:

40.6789km

243.098km

Explanation:

8 0
3 years ago
Two motorcyclists are riding side-by-side at night and the distance between their center-mounted headlights is 1.40 m. (a) If th
dezoksy [38]

Answer:

θ = 1.591 10⁻² rad

Explanation:

For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source

         

Let's write the diffraction equation for a slit

       a sin θ = m λ

The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ

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In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.

           θ = 1.22 λ / D

Where D is the diameter of the pupil

 Let's apply this equation to our case

        θ = 1.22 600 10⁻⁹ / 0.460 10⁻²

        θ = 1.591 10⁻² rad

This is the angle separation to solve the two light sources

6 0
3 years ago
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