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Vlada [557]
2 years ago
11

After fertilization in the fallopian tube, how many days will the zygote travel before arriving at uterus?.

Physics
1 answer:
natta225 [31]2 years ago
6 0

After fertilization in the fallopian tube, It will take 6-12 days for the zygote to travel before arriving at the uterus.

<h3>What is a zygote?</h3>

A zygote is, generally speaking, a cell created by the fusion of two gametes; the growing person is created from such a cell.

It takes the zygote around 6–12 days following fertilization in the fallopian tube for the fertilized egg to travel to the uterus and attach to the uterus, a process known as implantation.

Hence It will take 6-12 days for the zygote to travel before arriving at the uterus.

To learn more about the zygote refer;

brainly.com/question/465851

#SPJ1

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In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

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This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
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An object in circular motion has velocity that is constantly changing. The direction of the acceleration is
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<h2>Answer: Toward the center of the circle.</h2>

This situation is characteristic of the uniform circular motion , in which the movement of a body describes a circumference of a given radius with constant speed.  

However, in this movement the velocity has a constant magnitude, but its direction varies continuously.

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A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a
gizmo_the_mogwai [7]

Answer:

Acceleration,  767.08\ m/s^2

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

v^2=u^2+2ah

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a = acceleration=g

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2

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