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3 years ago

How does the resulting cell at the end of asexual reproduction compare to the original cell

Physics

1 answer:

lina2011 [118]3 years ago

4 0

1. At the end of asexual reproduction the resulting cell is an exact genetic copy of the original cell.

2. During inter phase the cell copies its DNA, and grows to an appropriate size where it is safe for the cell to split.

3. After meiosis, the resulting daughter cells are 4 genetically different haploid daughter cells.

4. During sexual reproduction, two haploid gametes join in the process of fertilization to produce a diploid zygote. The two haploid gamete bring different traits together, they also shares some DNA. This makes each daughter cell unique.

You might be interested in

Is magnesium a metal ?

tatuchka [14]

Yes Magnesium Its Metal

8 0

3 years ago

What would be the best design for an experiment that tests how much water expands when frozen?

galben [10]

B. Purchase a small plastic container and mark 1-ounce increments on the outside to determine volume. Pour 5 ounces of water into the container, and place in the freezer for 8 hours. Compare the frozen or ending volume with the liquid or beginning volume.

<h3>How much water expands when frozen?</h3>

Ice is less denser than the liquid form. Water is the only known non-metallic substance that expands when it freezes because it is the unique property of water. Water density decreases and it expands approximately about 9% by volume. For calculating the expansion of water, plastic container is the best option. We know that water expands when the water freezes because it is a unique property of water which allows the survival of aquatic organisms.

So we can conclude that option B is the right answer.

Learn more about water here: brainly.com/question/1313076

#SPJ1

5 0

2 years ago

A rock is tossed straight up from the ground with a speed of 21 m/s . When it returns, it falls into a hole 10 m deep.a.) What i

Arte-miy333 [17]

(a) 25.2 m/s

Let's take the initial vertical position of the rock as "zero" (reference height).

According to the law of conservation of energy, the speed of the rock as it reaches again the position "zero" after being thrown upwards is equal to the initial speed of the rock, 21 m/s (in fact, if there is no air resistance, no energy can be lost during the motion; and since the kinetic energy depends only on the speed of the rock:

$K=\frac{1}{2}mv^2$

and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)

This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using

$v^2 = u^2 + 2gd$

where

u = 21 m/s is the initial speed of the rock as it enters the hole

g = 9.8 m/s^2 is the acceleration due to gravity

d = 10 m is the depth of the hole

Substituting,

$v=\sqrt{u^2 +2gd}=\sqrt{(21 m/s)^2+2(9.8 m/s^2)(10 m)}=25.2 m/s$

(b) 4.72 s

The vertical position of the rock at time t is given by

$y(t) = v_y t - \frac{1}{2}gt^2$

where

$v_y = 21 m/s$ is the initial vertical velocity

Substituting y(t)=-10 m, we can then solve the equation for t to find the time at which the rock reaches the bottom of the hole:

$-10 = 21 t - \frac{1}{2}(9.8)t^2\\10+21 t -4.9t^2 = 0$

which has two solutions:

t = -0.43 s --> negative, so we discard it

t = 4.72 s --> this is our solution

7 0

4 years ago

Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0

BaLLatris [955]

Answer:

Part a)

$r = 6.96 \times 10^6 m$

Part b)

$F_g = 3.004 \times 10^3 N$

Part c)

$a = 0.88 m/s^2$

Part d)

$v = \frac{2480}{2} = 1240 m/s$

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

$v = 2480 m/s$

now we will have

$v = \sqrt{\frac{GM_{mars}}{r}}$

now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

$a = \frac{F_g}{m}$

$a = \frac{3004}{3400}$

$a = 0.88 m/s^2$

Part d)

As we know by III law of kepler

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

here we know that T2 = 8 T1

$(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}$

$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

as we know that speed is given as

$v = \sqrt{\frac{GM}{r}}$

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

$v = \frac{2480}{2} = 1240 m/s$

7 0

4 years ago

An ostrich can run at speeds of up to 72 kilometers per hour. How long will it take an ostrich to run 1.5 kilometers at this top

tatiyna

S=d/t
T=d/s
= 1.5/72
= 0.02 hours

6 0

4 years ago