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3 years ago

What is asexual propagation? (simple answer)

2 answers:

7 0

Asexual propagation involves taking a part of one parent plant and causing it to regenerate itself into a new plant. The resulting new plant is genetically identical its parent. Asexual propagation involves the vegetative parts of a plant: stems, roots, or leaves.

The most common method for asexual propagation is from cuttings. A cutting is any detached plant part which, under favorable conditions for regeneration of missing parts, will produce a new plant identical to the parent plant.

Hope that help:)

Firdavs [7]3 years ago

6 0

Asexual propagation is a type of plant reproduction using roots, stems, or leaves of a parent plant. Asexual propagation produces a genetically identical plant to the parent plant.

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A man inside the water seea a fish in water. will the fish appear to man in it's actual position? will the man appear to fish in

weqwewe [10]

Answer:

Explanation:

due to the refraction of light,meaning bending of light when it passes from diffrent mediums with diffrent densities

(probably mot the complete ans but smtg like this)

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3 years ago

Read 2 more answers

An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane

lubasha [3.4K]

Answer:

$\rho = 12580.7 kg/m^3$

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

$F = \frac{GMm}{(r + h)^2}$

here we have

$F =\frac {mv^2}{(r+ h)}$

$\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}$

here we have

$v = \sqrt{\frac{GM}{(r + h)}}$

now we can find time period as

$T = \frac{2\pi (r + h)}{v}$

$T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}$

$1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}$

$M = 4.54 \times 10^{23} kg$

Now the density is given as

$\rho = \frac{M}{\frac{4}{3}\pi r^3}$

$\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}$

$\rho = 12580.7 kg/m^3$

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2 years ago

You need to build a prototype with machined parts that withstand a saline corrosive environment and temperatures above 200 degre

kondor19780726 [428]

Answer:

Stainless steel

Explanation:

I will try to order the solutions from the least correct to the most correct.

Since a temperature greater than 200 ° F is required, that is to say approximately 93 ° c, Polycaprolactone is the least indicated. Its melting point is approximately 60 ° C, so it would not serve the required application.

On the other hand we have Untreated aluminum, which although it has a melting point higher than the required one, without a zinc and magnesium treatment it will easily oxidize in a salty environment, so it cannot be used in this choice either.

We have to compare the two steels.

The Mild Steel has a better corrosion resistance than the previous ones, but in a long-term cycle it will end up full of corrosion and therefore its properties will be highly affected.

Finally, we have stainless steel, which, as the name implies, contains in some of its variations chromium, zinc or magnesium in its alloys, which makes it highly resistant to corrosion.

In addition its melting point is above 1500 ° c.

The best choice is stainless steel.

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