A woman has one damaged fallopian tube. The damage completely blocks the opening of the tube at the ovary. How will this most likely affect her fertility?
Her likelihood of conceiving will be reduced by at least 50%.
Her eggs, if fertilized, will not implant properly.
Her eggs will not mature inside the ovary.
Her chances of conceiving twins will be doubled.

Answers

Answer 1

Answer:Her likelihood of conceiving will be reduced by at least 50%.

Explanation:

EDGE 2020

Answer 2

Answer:

Her likelihood of conceiving will be reduced by at least 50%

Explanation:

took the test on edge, got it right


Related Questions

A wave has a frequency of 30Hz and wave length of 40cm. What is the velocity of the wave?

Answers

Answer:

12m/s

Explanation:

v=fλ

30×(40÷100)=

12m/s

1 kg block slides down a frictionless inclined plane that makes an angle of 300 with respect to the ground. The total length of the plane is 2 m, but midway down it collides with a second block, weighing 0.5 kg. The two blocks stick together and travel as one unit the rest of the way down the ramp. What is the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane

Answers

Answer:

the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J

Explanation:

Given that the data in the question;

angle of inclination with respect to the ground [tex]\theta[/tex] = 30°

length of plane d = 2m

m₁ = 1 kg

m₂ = 0.5 kg

now, velocity of the first block at midpoint;

[tex]\frac{1}{2}[/tex]mv² = mgsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]

[tex]\frac{1}{2}[/tex]v² = gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]

v² = gsin[tex]\theta[/tex]d

v = √( gsin[tex]\theta[/tex]d)

g is 9.8 m/s

so we substitute

v = √( 9.8 × sin30° × 2)

v = √( 19.6 )

v =  3.13 m/s

Now, velocity just after collision of the blocks will be;

(m₁ + m₂)v₂ = m₁v

v₂ = m₁v / (m₁ + m₂)

we substitute

v₂ = (1 × 3.13) / (1 + 0.5)

v₂ = 3.13 / 1.5

v₂ = 2.0866 m/s

now, final kinetic energy will be;

[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + Initial Kinetic energy

[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + [tex]\frac{1}{2}[/tex]mv₂²

we substitute

[tex]KE_f[/tex] = [(1 + 0.5)9.8 × sin30 × [tex]\frac{2}{2}[/tex]] + [[tex]\frac{1}{2}[/tex] × 1.5 × 2.0866 ]

[tex]KE_f[/tex] = 7.35 + 3.2654

[tex]KE_f[/tex] = 10.62 J

Therefore, the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J

: A small block with mass 0.130 kg is attached to a string passing through a hole in a frictionless, horizontal surface. The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks

Answers

Answer:

r = 0.0173 m = 1.73 cm

Explanation:

Here, the centripetal force of the block will be providing the required breaking tension in the string:

[tex]Tension = Centripetal Force\\T = F_c\\\\T = \frac{mv^2}{r} \\\\r = \frac{mv^2}{T}\\[/tex]

where,

r = radius = ?

m = mass of block = 0.13 kg

v = tangential spee of block = 4 m/s

T = Breaking Strength = 30 N

Therefore,

[tex]r = \frac{(0.13\ kg)(4\ m/s)^2}{30\ N}[/tex]

r = 0.0173 m = 1.73 cm

When finding the radius of the string at the point it breaks, the tangential

velocity is assumed to be constant.

The radius when the string breaks is [tex]\underline{6.9 . \overline 3 \times 10^{-3}} \ m[/tex]

Reasons:

The mass of the small block, m = 0.130 kg

Initial radius of the circle of rotation = 0.800 m

Tangential velocity, v = 4.00 m/s

The radius of the path of rotation is reduced as the string is pulled

Breaking strength of the string = 30.0 N

Required:

The radius of the circle when the string brakes

Solution:

[tex]Centripetal \ force = \dfrac{m \cdot v^2}{r}[/tex]

Where;

r = The radius of the circle of rotation

When the string brakes, w have;

Centripetal force = Breaking strength of the string = 30.0 N

Which gives;

[tex]\displaystyle r = \mathbf{\dfrac{m \cdot v^2}{Centrifugal \ force}} = \frac{0.130 \times 4^2}{30} =6.9\overline 3 \times 10^{-2}[/tex]

The radius of the circle when, the string breaks r = [tex]\underline{6.9\overline 3 \times 10^{-2}} \ m[/tex]

Learn more here:

https://brainly.com/question/20905151


Give the relationship between the number of valence electrons in an atom's
valence electron shell and the position of the element on the Periodic Table​

Answers

Answer:

they're reactions

Explanation:

The relationship between the valence electrons and position is: the number of valence electrons determines the position

What is valence electron?

This is the number of electrons in the outermost shell of an atom.

NOTE: The outermost shell is called valence shell

Position in Periodic table

This is where an element is located in the periodic table

Relationship between valence electrons and position

The position of an element in the periodic table is determined by the number of valence electrons.

For example

Sodium, Na (atomic number of 11) has the following electronic configuration

1st shell = 2 electrons2nd shell = 8 electrons 3rd (valence) shell = 1 electron

Since the valence electron is 1, thus, sodium is located in group 1 of the periodic table.

Thus, we can see that the position of an element in the periodic table is related to the valence electron(s) in the atomic shell of the element.

Learn more about valence electron:

https://brainly.com/question/13993867

#SPJ2

Please Help!!!

For A&B

Answers

Answer:

A. 40N

B. 5m/s

Explanation:

A.

Impulse is equal to the area under the curve of a force vs. time graph. In this case, the area is in the shape of a triangle with base 8 (12-4=8) and perpendicular height 10:

Area of a triangle = (1/2)bh

A=(1/2)*8*10

=40

ANSWER: 40N

B.

Impulse = mass * velocity

40 = 8v

v = 5

ANSWER: 5m/s

Why do astronomers use frequencies other than the visible ones when they are
investigating the universe?

Answers

Because not all of the universe can be seen with a visible spectrum

Philosophy: The Big Picture Unit 8

Would an existentialist argue that the study of philosophy was a good use of a life?

A. Yes, if society valued the results of the study.
B. Yes, but only if the individual found it meaningful.
C. No, a life should be spent minimizing the role of anxiety.
D. No, there is no way that philosophy could create a meaningful life.

Answers

the answer to this would be C

1.A body of mass 10kg falling freely was found to be falling at a rat of 20m/s what force will stop the body in 2second?​

Answers

Answer:

50N

Explanation:

force it is falling with can be found by mass into acceleration and then devide by half to find force that could stop it in 2 sec

Before we make measurements, let's make sure we understand the circuit. 1. Select all of the following that correctly describe what a volt meter and ammeter measure. Select all that apply: A volt meter measures the potential difference (or voltage) across a circuit element. A volt meter measures the potential difference (or voltage) passing through a circuit element. A ammeter measures the electric current passing through a circuit element. A ammeter measures the electric current across a circuit element.

Answers

Answer:

the correct answers are a and c

Explanation:

In an electrical circuit there are two important quantities to measure, such as voltage and current.

Voltage is the potential difference between two points in a circuit

current is the number of electrons you pass through a given point per unit of time.

Now let's analyze each answer

a) true. The potential difference across an element

b) False. The potential difference is u field there is no physical entity that moves

c) True. The current is electrons in motion and these pass through the given element

d) False. There is a physical quantity that passes through the point

the correct answers are a and c

A house is lifted from its foundations onto a truck for relocation. The house is pulled upward by a net force of 2850 N. This force causes the house to move from rest to an upward speed of 15 cm/s in 5.0 s. What is the mass of the house?​

Answers

Answer:

m = 95000 kg

Explanation:

Given that,

Net force acting on the house, F = 2850 N

Initial speed, u = 0

Final speed, v = 15 cm/s = 0.15 m/s

We need to find the mass of the house. Let the mass be m. We know that the net force is given by :

F = ma

Where

a is the acceleration of the house.

So,

[tex]F=m\dfrac{v-u}{t}\\\\m=\dfrac{Ft}{(v-u)}\\\\m=\dfrac{2850\times 5}{(0.15-0)}\\\\m=95000\ kg[/tex]

So, the mass of the house is equal to 95000 kg.

A 5kg block rests on a 30° incline. The coefficient of static friction between the block and the incline is 0.20. How large a horizontal force must push on the block if the block is to be on the verge of sliding. a) up the incline, b) down the incline ? ​

Answers

Answer:

Hope It Help

Explanation:

That's all I know

sinat
Accelerationa
2 2.84
7 34
TABLE
in Elination
t2 t3 T2 2
1=0.04
2.29 1.25 1.28 1.271.61
2 460 = 0.00 4.59 1.16 1.081.12 1-25
3 so = 0.12 6.89 0.88
097 0.53
4. = 0.16 9.210.8
9.21 0.850.796. 82/0.67
(So
0.72 0.77 0.75 l 0-56/
0.28
49. al
27. 49
13 11%.
41. 2L
= 0.2 11.54​

Answers

Answer:

so you have a question

Explanation:

either way,you have a nice day

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