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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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The block has a weight of 75 lblb and rests on the floor for which μkμk = 0.4. The motor draws in the cable at a constant rate o
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The given question is incomplete. The complete question is as follows.

The block has a weight of 75 lb and rests on the floor for which \mu k = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.

Determine the output of the motor at the instant \theta = 30^{o}.

Explanation:

We will consider that equilibrium condition in vertical direction is as follows.

           \sum F_{y} = 0

         N - W = 0

           N = W

or,      N = 75 lb

Again, equilibrium condition in the vertical direction is  as follows.

        \sum F_{x} = 0

       T_{2} - F_{k} = 0

         T_{2} = \mu_{k} N

                  = 0.4 \times 75 lb

                  = 30 lb

Now, the equilibrium equation in the horizontal direction is as follows.

         \sum F_{x} = 0

       T Cos (30^{o}) + T Cos (30^{o}) = T_{2}

          2T Cos (30^{o}) = T_{2}

    or,             T = \frac{T_{2}}{2 Cos (30^{o})}

                        = \frac{30}{2 Cos (30^{o})}

                        = \frac{30}{1.732}

                        = 17.32 lb

Now, we will calculate the output power of the motor as follows.

             P = Tv

                = 17.32 lb \times 6

                = 103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}

                = 0.189 hp

or,             = 0.2 hp

Thus, we can conclude that output of the given motor is 0.2 hp.

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