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yan [13]
3 years ago
6

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:

D

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

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vova2212 [387]

Answer:

X2 is fasteer

x=0 will go to Xi

Explanation:

4 0
3 years ago
7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
artcher [175]

Answer:

20.4 m/s^{2}

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

7 0
3 years ago
A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear
Talja [164]

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

f_o = f_s\frac {(v+vs)}{v} and f_o = f_s\frac {(v-vs)}{v} when moving towards observer

With f_o of 76 then taking speed in air as 343 m/s we have

76 = f_s\times\frac {(343-vs)}{343}

f_s=\frac {343\times 76}{343-v_s}

Similarly, with f_o of 65 we have

65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}

Now

f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}

v_s=27.76 m/s

Substituting the above into  any of the first two equations then we obtain

f_s=70.07 Hz

4 0
3 years ago
A charge q produces an electric field of strength 2E at a distance of d away. Determine the electric field strength at a distanc
Alja [10]

The electric field strength of a point charge is inversely proportional to the square of the distance from the charge ... a lot like gravity.

If the magnitude of the field is (2E) at the distance 'd', then at the distance '2d', it'll be (2E)/(2²).  That's (2E)/4 = 0.5E .

3 0
3 years ago
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Murljashka [212]
You multiply the mass by the acceleration 82*7.5=615; that's what I would do
5 0
3 years ago
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