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faltersainse [42]
3 years ago
9

Why would most marine organisms live in photic zone?

Biology
1 answer:
BabaBlast [244]3 years ago
7 0
Upwelling carries nutrients from the deep waters into the photic zone, strengthening phytoplankton growth. ... Besides phytoplankton, many other animals also live in this zone and utilize these nutrients. The majority of ocean life occurs in the photic zone, the smallest ocean zone by water volume.
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How do materials move between cells ?​
DIA [1.3K]
<h2>Answer:</h2><h2>☠☠☠⭐⭐</h2>

The efficiency of movement of substances in and out of a cell is determined by its volume to surface area ratio.

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2 years ago
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What were pregnant women exposed to and what were some of the ways it affected their
Hoochie [10]

Answer:

Here is the answer and some! I hope it helps.

What were pregnant women exposed to and what were some of the ways it affected their children?

Teflon and they had deformities

2.What is the active ingredient in Teflon?

Fluorine

3.When the rats were exposed to c-8 what were some of the effects?

Developed cancers

Explanation:

5 0
3 years ago
5)
STatiana [176]

Answer:

D) the cell appears to be in a rectangular shape and has a cell wall

Explanation:

animal cells don't have cell walls

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3 years ago
Which scientist developed the idea that microorganisms can cause diseases
Vaselesa [24]

Answer:  Louis Pasteur

Louis Pasteur in 1860s, proposed germ theory. The theory was based upon the fact that diseases are caused by the microorganisms.He was officially the first to publish his work.

6 0
3 years ago
Gold has a specific heat of 0.129 J/(g×°C). How many joules of heat energy are required to raise the temperature of 15 grams of
Elodia [21]

Answer:

\boxed {\boxed {\sf 121.905 \ J  }}

Explanation:

We are asked to find the energy given mass, specific heat, and change in temperature. Therefore, we must use this formula;

q= mc \Delta T

The mass is 15 grams and the specific heat is 0.129 J/(g×°C). Let's calculate the change in temperature.

  • ΔT= final temperature - initial temperature
  • ΔT= 85 °C- 22°C = 63°C

Now we know all the values:

m= 15 \ g \ \\c= 0.129 \ J / (g* \textdegree C) \\\Delta T= 63 \ \textdegree C

Substitue the values into the formula.

q= (15 \ g)( 0.129 \ J / (g* \textdegree C)) ( 63 \ \textdegree C)

Multiply the first numbers together. The grams will cancel.

q= (1.935 \ J/ \textdegree C) ( 63 \ \textdegree C)

Multiply again, this time the degrees Celsius cancels.

q= 121.905 \ J

<u>121.905 Joules</u> are required.

5 0
3 years ago
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