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alisha [4.7K]
3 years ago
6

Which of these is NOT an interaction between a plant's root and shoot systems?

Biology
1 answer:
olga_2 [115]3 years ago
4 0

Answer:

The stomata open to allow carbon dioxide to enter the leaves.

Explanation:

Higher plants are made up of two major system parts namely: shoot system and root system. The shoot system constitutes every part of the plant above the ground i.e leaves, fruits, stem etc. while the root system constitutes the part below ground i.e the roots. The shoot system and root system, however, interacts in several ways to ensure that the plant functions optimally. The ways they interact include:

- The roots (root system) absorb water from the soil, which is transported upwards to prevent leaves (shoot system) from drying up.

- The phloem, which is a conducting tissue, transports sugars (product of photosynthesis) from the leaves (shoot system) to the root nodules (root system) for storage.

- Nutrients are absorbed from the soil by the roots (root system) and moved to the stems (shoot system).

Note that, although it is true that "stomata in the leaves open to allow carbon dioxide to enter the leaves", there is no connection between the shoot and root system.

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2.86 A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight
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Answer:

Distance Below the top = 6,02 m

Explanation:

To get the Final velocity of the first ball (that will be the intial velocity of the second) you need to solve the kinematic equation for Velocity:

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As the ball is dropped from rest V_{1O} = 0, so:

(2) V_{1f} = - gt

Note that the velocity is going to be negative as the ball is going down. To get the time it would take the ball to reach de base you can use the kinematic equation for position:

(3) h_{1} = h_{1O} + V_{1O} *t - \frac{1}{2} gt^{2}

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(4) 24m = \frac{1}{2} gt_{1f}^{2}

Solving for t, with 9,8\frac{m}{s^{2}} and :

(5) t_{1f} = \sqrt{\frac{24m*2}{g} } = 2,21 s

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(6) V_{1f} = -21,66  \frac{m}{s}

So the initial velocity of ball 2 is equial to this but oppossite in direction so: V_{2O}= -V_{1f} = 21,66  \frac{m}{s}

The general position equation for ball 2 is (considering h_{2O} = 0 :

(7) h_{2} = V_{2O} *t - \frac{1}{2} gt^{2}

They cross paths when h_{1}=h_{2} so:

(8) h_{1O} - \frac{1}{2} gt^{2} = V_{2O} *t - \frac{1}{2} gt^{2}

Rearranging:

(9) t_{cross} =\frac{h_{1O}}{V_{2O} }

Replacing values:

t_{cross} = 1,108 s

To get the absolute position replace t_{cross} on equation (3) or (7):

h_{cross} = 17,98 m

To get it below the top of te cliff:

h_{cross, from above} = 24 m - 17,98 m

h_{cross, from above} = 6,02 m

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Answer:

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