All categories

3 years ago

Which description properly describes the plant structures involved in photosynthesis?

Physics

2 answers:

nignag [31]3 years ago

7 0

Answer:

D-Stomata takes in Carbon Dioxide and release oxygen.

Explanation:

The answer is D because...

Photosynthesis rule. Photosynthesis takes in sunlight or carbon dioxide and releases oxygen.

Step one let's look at A Stomata takes in water, sunlight, and carbon dioxide and release oxygen. The reason this is not the answer is because taking in water is not a step in photosynthesis. Photosynthesis is ONLY sunlight or carbon dioxide.
Step two now we shall look at B , Phloem transports water, stomata take in carbon dioxide, and chlorophyll absorbs sunlight. This is not correct because once again water is not a step of photosynthesis and the phloem is NOT apart of the plant in which photosynthesis takes place. Plus the phloem does not transport water. The Xylem does.
Step three let's look at C, Xylem takes in water, sunlight, and carbon dioxide and releases oxygen. This is incorrect because once again photosynthesis does NOT require water or the Xylem. Xylem is the way water transports through out the plant.
Finally Step four. let's look at D. Stomata take in carbon dioxide and release oxygen. This is true because the Stomata acts like a gas change or a door. It takes in the carbon dioxide and the pores open to release oxygen for use mammals to breathe.
Hope this Helped :)

raketka [301]3 years ago

3 0

Answer:

Explanation:

Stomata take in carbon dioxide (i think they realease oxygen too), Phloem transports the glucose throughout the plant, Xylem transports the water and minerals from the roots. If you want me to explain more I can

You might be interested in

What's the most often given a value of zero to describe an object position on a straight line

Yanka [14]

B. The reference point is the 0 of a straight line description. Left of it would be negative and right of it would be positive.

8 0

3 years ago

Read 2 more answers

Which statement is true about a polarized object?

agasfer [191]

A polar substance has regions that are positive and negative, but there are no electrons exchanged. so the best answer is 5

8 0

3 years ago

Read 2 more answers

Using the superposition method, calculate the current through R5 in Figure 8-71

Natasha2012 [34]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

$r_1 = \frac{2.2 * 1}{2.2 + 1}$

$r_1 = 0.6875 k ohm$

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

$r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}$

$r_2 = 0.95 k ohm$

now the current flowing through the battery is

$i = \frac{2}{1 + 0.95} = 1.02 mA$

now this will divide into R3 and R2 so current flowing in R3 will be

$i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA$

now this will again divide in R4 and R5

so current in R5 will be

$i_5 = \frac{R_4}{R_4 + R_5}* i_1$

$i_5 = 0.18 mA$

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

$r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm$

also after its series with R3

$r_2 = 1 + 0.6875 = 1.6875 k ohm$

now it is in parallel with R5 on other side

$r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm$

now current through the battery will be given as

$i = \frac{3}{1 + 0.95} = 1.53 mA$

now it is divide in r2 and R5

so current in R5 is given as

$i_5 = \frac{r_2}{r_2 + R_5}*i$

$i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53$

$i_5 = 0.67 mA$

now the total current in R5 will be given by super position which is

$i = 0.67 + 0.18 = 0.85 mA$

so there is 0.85 mA current through R5 resistance

8 0

3 years ago

A snowball is launched horizontally from the top of a building at v = 16.9 m/s. If it lands d = 44 meters from the bottom, how h

Aloiza [94]

$v_x=16,9\frac{m}{s}\\
s=44m\\
g=10\frac{m}{s^2}\\\\
v_x=\frac{s}{t}\Rightarrow t=\frac{s}{v_x}\\\\
t=\frac{44}{16,9}\\\\
t=2,6s\\\\
h=\frac{gt^2}{2}\\\\
h=\frac{10*2,6^2}{2}\\\\
h=5*6,76\\\\
h=33,8m$