The retina of the eye does all of the following except _________

In some plants, the pistils don’t Form until a few days after the stamens do.How might this keep a plant from self pollinating

KATRIN_1 [288]

For a flower to be pollinated, pollen from an anther (which is located at the top of the stamen) needs to reach a stigma (at the top of the pistle.) Some plants are genetically capable of pollinating themselves if their own pollen reaches their own stigma; some plants are not capable of self pollination under any circumstances.

For plants that can genetically self pollinate, but would prefer not to, they can avoid this by having their pistil and pollen/stamens mature at different times. If the stamens mature first, the pollen will be dispersed by animals or wind or whatever dispersal mechanism it relies on. Then by the time the pistil is ready to be pollinated, there is no pollen left in that flower to land on the stigma.

3 0

3 years ago

Need help with this<br> THANKS TO WHO HELPS <3

nekit [7.7K]

Answer:

because it's north

Explanation:

u if yyggggggggggggggg hgfth

8 0

3 years ago

Read 2 more answers

Feng and Isaac are riding on a merry-ground. Feng rides on a horse at the outer rim of the circular platform, twice as far from

kogti [31]

Answer: The question is incomplete or missing details. here is the remaining part of the question ;

1. impossible to determine

2. half of Isaac’s

3. the same as Isaac’s

4. twice Isaac’s

The angular speed of feng will be the same as that of Isaac. Hence the answer is option 3

Explanation:

Since we have been told that both feng and isaac are riding on a merry go round i.e in a circular motion, irrespective of how fast one ride above the other, the angular speed will be constant since they are riding on a merry go round, as such both feng and isaac will maintain equal angular speed, hence the angular speed of feng will be the same as that of Isaac.

4 0

3 years ago

Why does a ball accelerate as it rolls down a hill?

Anika [276]

The ball accelerates because of gravity.

6 0

3 years ago

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A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

So

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$