What is the end product of mitosis?
A. Four gametes
B. Two identical nuclei
C. One complete zygote
D. A cell with one set of chromosomes

What Is The End Product Of Mitosis?A. Four GametesB. Two Identical NucleiC. One Complete ZygoteD. A Cell

Answers

Answer 1

Answer:

It is B

Explanation:

I did the test

Answer 2

From the given options, the end product of mitosis is Two identical nuclei (option b)

Mitosis is a fundamental cellular process that occurs in somatic (non-reproductive) cells, aiming to create two genetically identical daughter cells from one parent cell. It consists of several stages: prophase, metaphase, anaphase, and telophase.

During prophase, the chromatin (a combination of DNA and proteins) condenses into visible chromosomes, and the nuclear membrane disintegrates. In the subsequent stage, metaphase, the chromosomes align at the cell's equator, ensuring each daughter cell receives an equal number of chromosomes.

Next comes anaphase, where the paired chromosomes are pulled apart and move towards opposite poles of the cell. The final phase, telophase, involves the formation of two separate nuclei around the separated chromosomes.

At the end of mitosis, the single parent cell has divided into two genetically identical daughter cells, each containing a complete set of chromosomes. These daughter cells are fully functional and capable of carrying out their specific tasks within the organism.

So, the correct Answer: B. Two identical nuclei.

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Related Questions

8
If you put 100 carbon atoms in a box. How many carbon atoms remain after 2 half-lives or
11460 years? (you might need to review half-lives to answer this question) *
(1 Point)
100 atoms
50 atoms
25 atoms
12 atoms

Answers

Answer:

25 atoms

Explanation:

Lily can run 100 m in 10s. How fast was her average speed?

Answers

Answer:

The answer is 10 m/s

Explanation:

The average speed can be found by using the formula

[tex]v = \frac{d}{t} \\ [/tex]

d is the distance

t is the time taken

From the question we have

[tex]v = \frac{100}{10} \\ [/tex]

We have the final answer as

10 m/s

Hope this helps you

A boy throws a steel ball straight up. Consider the motion of the ball only after it has left the boy's hand but before it touches the ground, and assume that forces exerted by the air are negligible. For these conditions, the force(s) acting on the ball is (are):
a) a downward force of gravity along with a steadily decreasing upward force.
b) a steadily decreasing upward force from the moment it leaves the boy's hand until it reaches its highest point; on the way down there is a steadily increasing downward force of gravity as the object gets closer to the earth.
c) an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity.
d) an almost constant downward force of gravity only.
e) none of the above. The ball falls back to the ground because of its natural tendency to rest on the surface of the earth.

Answers

Answer:

c) an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity.

Explanation:

When the boy throws the ball upward, there are two forces acting on the ball. First one is the force of gravity, which is constant and equal tot he weight of the ball, acting down. The second force is the force with which the boy throws the ball. This force acts upward and decreases as the ball moves up. This force becomes zero when the ball reaches highest point.

While, on its way down the ball has only one force acting on it. That force is the constant force of gravity acting downward. This gravity force is equal to the weight of the ball.

Hence, the correct option is:

c) an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity.

The force acting on the ball is an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity. Thus the option C is the correct option.

Given,

Force exerted by the air is negligible.

Now the forces acting on the ball after the realese,

Gravity- This force acts on the body and will remain constant until the ball is in the air. This force is equal to the weight ([tex]W[/tex]) which is equal to the product of the mass [tex]m[/tex] and the gravitational force [tex]g[/tex] of the earth.

            [tex]W=mg[/tex]

Force applied by the boy- This is the force by which the boy through the ball. This will act upward and keep on decreasing as the ball moves upward. This force becomes zero when the body reaches the highest point.

Now after reaching the highest point, the ball starts to come down. In this case, only one force will act on it which is the gravity force and this is equal to the weight of the body.

Hence, the force acting on the ball is an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity. Thus the option C is the correct option.

For more about the gravity follow the link below-

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For brainliest !!!!!!! Do it

Answers

Answer and Explanation:

Using a = F/m, we get:

Acceleration for Road A = 480/62 = ~7.74 m/s^2

Acceleration for Road B = 240/62 = ~3.87 m/s^2

Acceleration for Road C = 600/62 = ~9.68 m/s^2

If the net force becomes higher than 600N and the mass stays constant, the acceleration will increase proportionally to how much the force increases and the acceleration will be greater than the acceleration when the force is 600N.

I hope this helps! :)

A flagpole 24 meters high casts a shadow 33 meters long. What is the distance between the tip of the shadow and flagpole? Round your answer to the nearest meter

Answers

Answer:

The distance between the tip of the shadow and flagpole is 40.80 m.

Explanation:

Let the flagpole be represented by AB and the shadow BC . We have to find the distance between C and A .

Considering the right angle CA is the hypotenuse .

Using the Pythagoras theorem

(Hypotenuse)² = (base)² + (perpendicular)²

CA= √(24)² + (33)²

CA= √576+1089

CA= 40.80 m

The distance between the tip of the shadow and flagpole is 40.80 m.

Anyone play Among US?

Answers

Answer:

Yes

Explanation:

Its a good game.

Answer:

oof

Explanation:

A construction worker accidentally drops a brick from a high scaffold. a. What is the brick's velocity after 4.0 s? b. How far does the brick fall during this time?

Answers

Answer:

A. 39.2 m/s

B. 78.4 m

Explanation:

Data obtained from the question include:

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

A. Determination of the brick's velocity.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = gt

v = 4 × 9.8

v = 39.2 m/s

Thus, the brick's velocity after 4 s is 39.2 m/s

B. Determination of how far the brick fall in 4 s.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

h = ½gt²

h = ½ × 9.8 × 4²

h = 4.9 × 16

h = 78.4 m

Thus, the brick fall 78.4 m during the time.

The velocity after 4 seconds is 40 m/s and the height is 80 m.

From the equations of motion under gravity;

v = u + gt

V = final velocity of the brick

u = initial velocity of the brick

g = acceleration due to gravity = 10 ms-2

t = time taken

Note that, u = 0 m/s because the brick was dropped from a height.

v = 0 + 10(4)

v = 40 m/s

The height is obtained from;

h = ut + 1/2 gt^2

u = 0 m/s because the brick was dropped from a height.

h =  1/2 gt^2

h = 1/2 × 10 × (4)^2

h = 80 m

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The specific heat capacity of oxygen is 918 J/kg °C. The specific heat capacity of water is 4181 J/kg °C. Which substance will require more energy to raise its temperature by 1°C?

Oxygen and water

Answers

Answer:

Water

Explanation:

Answer:

So, water has a specific heat of exactly 1.00 calorie per gram per °C. Because one calorie is equal to 4.18 J, it takes 4.18 J to raise the temperature of one gram of water by 1°C. In joules, water's specific heat is 4.18 J per gram per °C.

The answer is water.

A teacher asks students to identify which demonstrations show physical changes and which show chemical changes.
Which demonstration shows a physical change?
A. Baking soda is mixed with vinegar. The mixture bubbles.
B. A dish of water is placed in the freezer. The water turns solid.
Ο Ο Ο Ο
C. A piece of pH paper is dipped in clear ammonia. The paper turns from yellow to pink.
D. Hydrogen peroxide solution is dropped on a potato slice. The liquid bubbles and fizzes.

Answers

Answer:

B. A dish of water is placed in the freezer and turns solid

Explanation:

it is a physical change because it can return to its original state by physical means. the other options are chemical changes because things like color change, bubbles or fizzing, and odor change all indicate a chemical change and cannot return to the original state it was in before.

18) Why do you think divergent plate boundaries and transform fault boundaries only occur at places where the crust is the same on both plates? ​

Answers

Answer:

A divergent boundary occurs when two tectonic plates move away from each other. Along these boundaries, earthquakes are common and magma (molten rock) rises from the Earth's mantle to the surface, solidifying to create new oceanic crust. When two plates come together, it is known as a convergent boundary.

Explanation:

17.
A car is accelerated at 4.0 m/s2 from rest. How long does it take the car to reach a speed of 28 meters per second?

Answers

Answer:

t = 7 seconds

Explanation:

Given that,

The acceleration of a car, a = 4 m/s²

The initial velocity of the car, u = 0

Final velocity of the car, v = 28 m/s

We need to find how much time the car to reach a speed of 28 m/s. Using the definition of the acceleration of a car. It is given by :

[tex]a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{28}{4}\\\\t=7\ s[/tex]

So, it will take 7 seconds to reach a speed of 28 m/s.

if a truck goes 30 kilometers in 30 minutes what is the average speed ​

Answers

Answer:

60 km per hour

Explanation:

if you drive for 30 min and go 30km then if you drive for 60 min(1hr) then you would have driven 60km.

Answer:

60

Explanation:

hope this helps

Two blocks are arranged at the ends of a mass-less string as shown in the figure. The system starts from rest. When the 3.26 kg mass has fallen through 0.386 m, its downward speed is 1.28 m/s. What is the frictional force between the 4.52 kg mass and the table? (round your answer to the 4th decimal place)

Answers

Answer:  002 10.0points Two blocks are arranged at the ends of a massless string as shown in . 96 kg mass has fallen through 0 . 415 m, its downward speed is 1 . 25 m / s.

Explanation: I hoped that helped!!

Answer:

2.5 kg

Explanation:

9. According to statistical theory, 68% of your measurements of time should fall within the range of and of therefore about 3 of your 5 measurements should satisfy this condition. How many of your measurements fall within this range? Are your errors random according to this theory?

Answers

Answer:

Yes

Explanation:

Yes, this is a random error generating because of statistical constraint. We only have finite number of data points. As per this, if we plot our observation we will get a gaussian (inverse bell ) shaped curve with mean equal to central value.

please help me.
During a football game, one of the players on the home team kicks the football that has a
mass of 0.6 kg so that the ball accelerates toward the opposing team at
23 m/s2. If no other forces act on the ball, how much force did the kicker apply to the
football?

Answers

Answer:

[tex]Force = 13.8N[/tex]

Explanation:

Given

[tex]Mass = 0.6kg[/tex]

[tex]Acceleration = 23m/s^2[/tex]

Required

Determine the applied force

From the question, we understand that no other force acts on the ball.

i.e. the only applied force on the ball is the force applied by the striker.

So, we apply Newton's second law to solve this question.

And this implies that:

[tex]Force = Mass * Acceleration[/tex]

[tex]Force = 0.6kg * 23m/s^2[/tex]

[tex]Force = 13.8N[/tex]

Hence, the applied force by the striker on the ball is 13.8N

Explain why people have different skin colors?

Answers

I think it because of UV rays ultra violet ray which can make their colors different

While playing football, Chris runs 4.5 m at 20⁰ south of west. If an opponent was trying to tackle him how far west and south would he have to run to reach him?

Answers

Answer:

South = 1.5m

West =4.2m

Explanation:

Kindly see attached a rough draft of the situation

Step one

Given data

From the sketch the direction of the player is along the resultant of the triangle, corresponding to the Hypotenuse

Step two:

Hence for an opponent to tackle him towards the south, he must be at

sin θ= opp/hyp

sin 20=x/4.5

x=sin 20*4.5

x=0.342*4.5

x= 1.5m

Also, for an opponent to tackle him towards the south, he must be at

cos θ= adj/hyp

cos 20=y/4.5

y=cos 20*4.5

y=0.93*4.5

y= 4.2m

If someone has a curve in his or her spine, what other bones might also be affected by the way the curve shifts that person’s weight?

Answers

Answer:

Curve in spine can effect his or her other body parts too in addition to spine. The neck or head can be tilted forward, shoulder could be hunched, hip bone can be uneven, ribs can be effected leading to disorder of lung.

Explanation:

the photos below show four pairs of objects. which pair of objects is experiencing the greatest gravitational force.A. the person and dog B. The person and the ball C. the person and earth D. the person and the building

Answers

Answer:

C.

Explanation:

the earth has a high greavitational pull

Answer: C the person and earth

Explanation: I just answered it on a p e x

PLEASE HELP !
What is the acceleration of a ball that starts at rest and increases in speed
to 20 m/s in 45 seconds? *

Answers

20/45 = 0.4? I’m not sure but I think this is it
20/45 which will be 0.4:))

which factor is needed to describe velocity but is not needed to describe speed

Answers

Answer:

Magnitude and direction describe velocity

However if there is only magnitude its mean it is speed.

Fun : probably you have seen speed movie. It could not have director because if speed is directed then it will be velocity lol..

When a bug hits a windshield of a car, the force of the car on the bug equals the force of the bug on the car.
O True
O False

Answers

True. The force of the car on the bug will be equal to the force of the bug on the car.

Newton's third law

This statement is true according to Newton's third law of motion.

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction.

In this case, the action is the force of the car on the bug, and the reaction is the force of the bug on the car.

Therefore, the force of the car on the bug is equal in magnitude and opposite in direction to the force of the bug on the car.

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Can y’all please help

Answers

Answer:

The cat's speed when it slid off the table was 2.53 m/s

Explanation:

Horizontal Motion

When an object is thrown horizontally at a speed v, from a height h, it describes a curved path ruled by gravity, until it hits the ground.

the range or maximum horizontal distance traveled by the object can be calculated as follows:

[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]

The situation described in the problem fits into the concept of a horizontal launch when the cat slides off the table and gets to the ground some distance ahead.

It's given the range as d=1.3m and the height h=1.3 m, thus we can find the cat's speed by solving the equation for v:

[tex]\displaystyle v=d\cdot\sqrt{\frac {g}{2h}}[/tex]

Where [tex]g=9.81\ m/s^2[/tex]

Substituting values:

[tex]\displaystyle v=1.3\cdot\sqrt{\frac {9.81}{2\cdot 1.3}}[/tex]

[tex]\displaystyle v=1.3\cdot\sqrt{\frac {9.81}{2.6}}[/tex]

Calculating:

v = 2.53 m/s

The cat's speed when it slid off the table was 2.53 m/s

(NEED HELP ASAP)
In a lab investigation one group of students (group a) measures the speed of a 0.1 toy car to be 2.5 m/s at the bottom of a ramp another group of students (group b) measures the speed of the car to be 3 m/s at the bottom of the ramp the cars starting position at the top of the ramp is 1 m high.

a. what is the potential energy of the car at the beginning of the experiment before it's speed is measured?

b. Calculate the kinetic energy of the car for group a.

C. Calculate the kinetic energy for the car for group b.

Answers

Answer:

a = 0.98J

Explanation:

PE = mgh

PE = 0.1 x 9.8 x 1

PE = 0.98J

(a) The potential energy of the car at the beginning of the experiment is 9.8(m) J.

(b) The kinetic energy of the car for group a is 3.125(m) J.

(c) The kinetic energy of the car for group b is 4.5(m) J.

Potential energy of the car at the beginning of the experiment

The potential energy of the car is calculated as follows;

P.E = mgh

where;

m is mass of the carg is accleration due to gravityh is height of the car

P.E = m x 9.8 x 1

P.E = 9.8(m) J

Kinetic energy of the car for group a

K.E = ¹/₂mv²

K.E = ¹/₂ x m x (2.5)²

K.E = 3.125(m) J

Kinetic energy of the car for group b

K.E = ¹/₂mv²

K.E = ¹/₂ x m x (3)²

K.E = 4.5(m) J

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A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is fired at an upward angle of 34.5° to the horizontal. Determine: (a) the time the ball is in the air: (b) the impact velocity: (c) the horizontal distance out from the base of the cliff that the ball strikes the water.​

Answers

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m​

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

[tex]t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085[/tex]

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ  and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m​

A sky diver jumps out of a plane and reaches terminal velocity at 91 meters per second. How far will she fall in 35 seconds?
Can I get this answer quick!

Answers

The answer is 3,185 meters

A researcher may use a two-tailed test to evaluate a directional hypothesis.
Please select the best answer from the choices provided
OT
OF

Answers

Answer:

true

Explanation:

just took the test

Answer:

True

Explanation:

Which is the length of the route between two points

Answers

Answer:

Can you show us the route and then I would be happy to help :)

Explanation:

A small plane flew 200 mph with a 40 mph tailwind for 480 miles. The plane returned home at a speed of 200 mph with a 40 mph headwind for 480 miles. How much longer did the flight home take?

Answers

Answer:

1 hour

Explanation:

The tailwind usually adds to the speed of airplanes while the headwind normally decreases the speed of airplanes.

For the initial journey of 200 mph with a 40 mph tailwind speed for 480 miles:

Total speed = 200 + 40 = 240 mph

Also,

Speed = distance/time

time = 480/240 = 2 hours

For the return journey of 200 mph with a 40 mph headwind for 480 miles:

Total speed = 200 - 40 = 160 mph

time = 480/160 = 3 hours

The initial flight took 2 hours while the flight back home took 3 hours. Hence it took 1 hour longer to get back home

* 1 po
What are the health benefits of having muscular strength and endurance?

Answers

Answer:

i can`t help without showing me the choses

Explanation:

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