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4 years ago

When pollen from one plant lands on the pistol of a flower on the same plant

Physics

1 answer:

bekas [8.4K]4 years ago

7 0

Answer:

Self-pollination occurs when pollen from one plant lands on the pistil of a flower on the same plant. Cross-pollination occurs when pollen from one plant reaches the pistil of a flower on a different plant.

Explanation:

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A river flows due south with a speed of 2.0 m/s .You steer a motorboat across the river; your velocity relative to the water is

mihalych1998 [28]

Answer:

a) $v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

b) $\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

c) $t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

d) $Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South

Explanation:

Part a

For this case the figure attached shows the illustration for the problem.

We know that $v_y = 2 m/s$ represent the velocity of the river to the south.

We have the velocity of the motorboard relative to the water and on this case is $V_x= 4.8 m/s$

And we want to find the velocity of the motord board relative to the Earth $v_m$

And we can find this velocity from the Pythagorean Theorem.

$v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

Part b

We can find the direction with the following formula:

$\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

Part c

For this case we can use the following definition

$D = Vt$

The distance would be D = w = 600 m and the velocity V = 4.8m/s and if we solve for t we got:

$t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

Part d

For this case we can use the same definition but now using the y compnent we have:

$Y = v_y t$

And replacing we got:

$Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South

7 0

3 years ago

Measurements of how orbital speeds depend on distance from the center of our galaxy tell us that stars in the outskirts of the g

diamong [38]

Answer:

True

Explanation:

Measurements of how orbital speeds depend on distance from the center of our galaxy tell us that stars in the outskirts of the galaxy. True or False

stars in the outskirt orbit the galactic center

the object mean orbital speed depends only on the Earth's mass and the semi-major axis (half the longest diameter) of the object's orbit. worthy of note. the orbital speed changes depending on where in the orbit the object is. It will be greatest when closest to Earth and least when furthest from Earth.

5 0

3 years ago

A/an select one is a type of electric circuit in which the overall resistance of the circuit increases as items are added to the

qwelly [4]

Parallel circuit<span>, as the number of resistors increases, the overall current also increases.</span>

4 0

3 years ago

A natural water molecule (H2O) in its vapor state has an electric dipole moment of magnitude, p = 6.2 x 10-30 C.m. (a) Find the

Vika [28.1K]

Answer:

a $D = 3.9 *10^{-12} \ m$

b $\tau_{max} = 1.24 *10^{-25} \ N\cdot m$

c $W = 2.48 *10^{-25} J$

Explanation:

From the question we are told that

The magnitude of electric dipole moment is $\sigma = 6.2 *10^{-30} \ C \cdot m$

The electric field is $E = 2*10^{4} \ N/C$

The distance between the positive and negative charge center is mathematically evaluated as

$D = \frac{\sigma }{10 e}$

Where e is the charge on one electron which has a constant value of $e = 1.60 *10^{-19} \ C$

Substituting values

$D = \frac{6.20 *10^{-30}}{10 * (1.60 *10^{-19})}$

$D = 3.9 *10^{-12} \ m$

The maximum torque is mathematically represented as

$\tau_{max} = \sigma * E * sin (\theta)$

Here $\theta = 90^o$

This because at maximum the molecule is perpendicular to the field

substituting values

$\tau_{max} = 6.2 *10^{-30} * 2*10^{4} sin ( 90)$

$\tau_{max} = 1.24 *10^{-25} \ N\cdot m$

The workdone is mathematically represented as

$W = V_{(180)} - V_{0}$

where $V_{(180)}$ is the potential energy at 180° which is mathematically evaluated as

$V_{(180) } = - \sigma * E cos (180)$

Where the negative signifies that it is acting against the field

substituting values

$V_{(180) } = - 6.20 *10^{-30} * 2.0 *10^{4} cos (180)$

$V_{(180) } = 1.24*10^{-25} J$

and

$V_{(0)}$ is the potential energy at 0° which is mathematically evaluated as

$V_{(0) } = - \sigma * E cos (0)$

substituting values

$V_{(0) } = - 6.20 *10^{-30} * 2.0 *10^{4} cos (0)$

$V_{(0) } =- 1.24*10^{-25} J$

So $W = 1.24 *10^{-25} - [-1.24 *10^{-25}]$

$W = 2.48 *10^{-25} J$

6 0

4 years ago

Describe how the motion of a planet (like Mars) among the stars of the zodiac on the sky is qualitatively different from the mot

Advocard [28]

Explanation:

In the plane of the Solar System, called Ecliptic plane, Sun is at the center and all the planets including Earth are going around the Sun in their respective orbits. As seen form the Earth, Sun appear to go around it and in this process it crosses through 12 constellations. These constellations are called as Zodiacs. Now since all the planets move in almost the same plane, they will appear in the sky to be lying near the ecliptic.

Sun's own motion is not that evident to us because of which the impact of that motion is nullified. But, motion of planet is clearly visible and it also shows in their annual round around the Sun. Here, Mars also shows retrograde motion which means it will show a back and forth motion in the sky. Because of which Mars might be visible in the same zodiac for a longer Duration as compared to the Sun.

7 0

3 years ago