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Andrei [34K]
3 years ago
9

Aquatic ecosystems consist of two different layers with respect to light. The photic zone is where light penetrates the water, a

t the surface and a little below. The aphotic zone is where the water is perpetually dark because the light cannot penetrate that deep into the water. The zone depths vary due to salinity and clarity of water.
Which of the following must be true about the organisms living in the aphotic zone?
A.
They have adapted to having no light for an energy source and are able to live in dark and cold areas.
B.
They all must move to the surface of the water once a day to receive adequate light energy.
C.
They have all become fluorescent so they can all be seen by one another.
D.
Many organisms have not been able to survive in the cold, dark waters, so population is low.
Physics
2 answers:
Lostsunrise [7]3 years ago
7 0

Answer: a

Explanation: it talks about how they live without that much light

Ivahew [28]3 years ago
4 0

Answer:

Answer A

Explanation:

Did the study island lesson

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There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is
Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

4 0
3 years ago
I’ll give brainliest!! please help and answer correctly! plsss answer quick
Rashid [163]

Answer: The motion of the object will remain the same

Explanation:

6 0
3 years ago
HELLO.
aivan3 [116]

Answer:

Hello! Thanks! I hope you are too! hope this helps!

Explanation:

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3 0
3 years ago
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A mail carrier leaves the post office and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00
den301095 [7]

Answer:

\theta=7^o

Explanation:

<u>Displacement</u>

It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.

The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is

\vec r_1=\ km

Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by

\vec r_2=\ km=\ km

Finally, he drives 9.5 km 35° north of east.

\vec r_3=\ km=\ km

The total displacement is

\vec r_t=\ km+\ km+\ km

\vec r_t=\ km

The direction can be calculated with

\displaystyle tan\theta=\frac{1.39}{11.28}=0.1232

\boxed{\theta=7^o}

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3 years ago
The box as shown above is moving at a constant velocity . True or false
STatiana [176]
True because on side is heavier than the other
8 0
3 years ago
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