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2 years ago

What's the difference between asexual reproduction and sexual reproduction?!?!

2 answers:

8 0

Hello!

This is simple, asexual reproduction involves only one parent, and does not need to involve the fusion of gametes. On the other hand, Sexual reproduction involves two parents and does need to fuse together different gametes and needs the genes from both sexes of the plants.

I hope this helped!

I am, yours most sincerely,
SuperHelperThingy

Leni [432]2 years ago

5 0

Sexual reproduction is the combination between the genetic material of both two parents while asexual reproduction produces offspring genetically identical to the one parent. Also, asexual reproduction produces without sex while sexual reproduction requires it. These are the most common differences between the two reproduction processes.

I hope this helps!

You might be interested in

Does melting change substances into other substances?

spayn [35]

Answer:

Yes, I think

Explanation:

Melting is a process that causes a substance to change from a solid to a liquid. Melting occurs when the molecules of a solid speed up enough that the motion overcomes the attractions so that the molecules can move past each other as a liquid.

4 0

2 years ago

Read 2 more answers

A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium a

nadya68 [22]

Answer:

82.25 moles of He

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 10 L

Mass of He = 0.329 Kg

Temperature (T) = 28.0 °C

Molar mass of He = 4 g/mol

Mole of He =?

Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

0.329 Kg = 0.329 Kg × 1000 g / 1 Kg

0.329 Kg = 329 g

Thus, 0.329 Kg is equivalent to 329 g.

Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:

Mass of He = 329 g

Molar mass of He = 4 g/mol

Mole of He =?

Mole = mass / molar mass

Mole of He = 329 / 4

Mole of He = 82.25 moles

Therefore, there are 82.25 moles of He in the tank.

8 0

2 years ago

Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the

zlopas [31]

Answer:

$\dfrac{I_1}{I_2}=\dfrac{4}{9}$

Explanation:

c = Speed of wave

$\rho$ = Density of medium

A = Area

$\nu$ = Frequency

$\nu_1=\dfrac{2}{3}\nu_2$

Intensity of sound is given by

$I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2$

So,

$I\propto \nu^2$

We get

$\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}$

The ratio is $\dfrac{I_1}{I_2}=\dfrac{4}{9}$

8 0

2 years ago

The 6 strings on a guitar all have about the same length and are stretched with about the same tension.The highest string vibrat

ahrayia [7]

Answer:

B. d(low)=4d(high)

Explanation:

Frequency of a string can be written as;

f = v/2L

Where;

v = sound velocity

L = string length

Frequency can be further expanded to;

f = v/2L = (1/2L)√(T/u) ......1

Where;

m= mass,

u = linear density of string,

T = tension

p = density of string material

A = cross sectional area of string

d = string diameter

u = m/L .......2

m = pAL = p(πd^2)L/4 (since Area = (πd^2)/4)

f = (1/2L)√(T/u) = (1/2L)√(T/(m/L))

f = (1/2L)√(T/((p(πd^2)L/4)/L))

f = (1/2L)√(4T/pπd^2)

f = (1/L)(1/d)√(4T/pπ)

Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.

f ~ 1/d

So, if

4f(low) = f(high)

Then,

d(low) = 4d(high)

6 0

3 years ago

A cylindrical, 0.500-m rod has a diameter of 0.02 m. The rod is stretched to a length of 0.501 m by a force of 3000 N. What is t

Salsk061 [2.6K]

Answer:

Y = 4.775 x 10⁹ Pa = 4.775 GPa

Explanation:

First, we calculate the stress on the rod:

$stress = \frac{Force}{Area} = \frac{3000\ N}{\pi r^2} \\\\stress = \frac{3000\ N}{\pi (0.01\ m)^2}\\\\stress = 9.55\ x\ 10^6\ Pa = 9.55 MPa\\$

Now, we calculate the strain:

$strain = \frac{Change\ in Length}{Original\ Length}\\\\strain = \frac{0.501\ m - 0.5\ m}{0.5\ m}\\\\strain = 0.002\\$

Now, we will calculate the Young's Modulus (Y):

$Y = \frac{stress}{strain}\\\\Y = \frac{9.55\ x\ 10^6\ Pa}{0.002} \\$

Y = 4.775 x 10⁹ Pa = 4.775 GPa

6 0

2 years ago