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Anika [276]
3 years ago
6

What must be true the number of chromosomes in each sex cell

Physics
1 answer:
zvonat [6]3 years ago
8 0

Answer:

In humans, each cell normally contains 23 pairs of chromosomes, for a total of 46.

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Describe liquids and GASSES IN TERMS OF THERE VOLUME AND SHAPES
antiseptic1488 [7]

Answer:

Explanation:

liquids have definite volume

liquids do not have definite shape. The take the shape of the container in which they are kept.

gases do not have definite volume.

gases do not have definite shape. They take the shape of the container in which they are kept.

Hope this helps

plz mark as barinliest!!!!!!

Stay safe!

3 0
3 years ago
Read 2 more answers
A charge of 90 C passes through a wire in 1 hour 15 minutes . what is the current in the wire
timurjin [86]
We calculate current from the formula:
I= \frac{q}{t} , where q is a electric charge transferred over time t 
Time should be converted to seconds:
1h 15 min= 75min= 4500s
I=\frac{90C}{4500s}=0,02A Result is in unit-Ampere
5 0
3 years ago
Read 2 more answers
An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
Zigmanuir [339]

Answer:

(A) therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

(B) therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

\frac{1}{f} =\frac{1}{s'} +\frac{1}{s} where

  • s' = distance of the image
  • f = focal length
  • but we first need to find the focal length before we can apply this formula

\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )

\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )

\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )

\frac{1}{f} =\frac{0.7}{12.5}

f = \frac{12.5}{0.7}

f = 17.9 cm

now that we have the focal length we can apply \frac{1}{f} =\frac{1}{s'} +\frac{1}{s}

\frac{1}{f} - \frac{1}{s} =\frac{1}{s'}

\frac{1}{17.9} - \frac{1}{25} =\frac{1}{s'}

\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}

\frac{7.1}{447.5} =\frac{1}{s'}

s' = \frac{447.5}{7.1}[/tex]  = 63 cm to the right of the lens

magnification =\frac{-s'}{s} =\frac{y'}{y}   where y' is the height of the image, therefore

\frac{-s'}{s} =\frac{y'}{y}

\frac{-63}{25} =\frac{y'}{52.5}

y' = \frac{-63}{25} x 0.525 = -13.22 cm

therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

\frac{1}{f} =\frac{1}{s'} +\frac{1}{s} where

  • s' = distance of the image
  • f = focal length
  • but we first need to find the focal length before we can apply this formula

\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )

\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )

\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)

\frac{1}{f} =\frac{0.7}{12.5}

f = \frac{12.5}{0.7}

f = 17.9 cm

now that we have the focal length we can apply \frac{1}{f} =\frac{1}{s'} +\frac{1}{s}

\frac{1}{f} - \frac{1}{s} =\frac{1}{s'}

\frac{1}{17.9} - \frac{1}{25} =\frac{1}{s'}

\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}

\frac{7.1}{447.5} =\frac{1}{s'}

s' = \frac{447.5}{7.1}[/tex]  = 63 cm to the right of the lens

magnification =\frac{-s'}{s} =\frac{y'}{y}   where y' is the height of the image, therefore

\frac{-s'}{s} =\frac{y'}{y}

\frac{-63}{25} =\frac{y'}{52.5}

y' = \frac{-63}{25} x 0.525 = -13.22 cm

therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

7 0
3 years ago
What has been gravity's apparent role in our solar system so far? Focus on the formation of planets, individual orbits, alignmen
masya89 [10]

Gravity is one of the significant forces of the universe. It is the force that draws other objects closer to the center of another object. The bigger the object, the higher its force of gravity and nearby objects are accelerated towards the big mass.

It is this force of gravity that keeps the planets in the solar system orbiting around the sun. The suns gravitational pull is stronger than those of the planets hence pulls the planets on their orbits around the star. Gravity also played a part in the formation of the planets. Gravity caused the condensation of dust and rocks into a mass that continually attracted more matter as it gained mass (due to gaining more associated gravitational pull).


6 0
3 years ago
When fireworks explode, sound and light are produced. These are examples of
ki77a [65]

Answer: Macroscoptic Output

Explanation:

Answers to the rest:

1. B) macroscopic outputs.

2.A) a microscopic change creating a macroscopic output

3.B) Because the energy levels of the electrons in different metals are usually not the same, different metals usually emit different colors of visible light.

4.A) Heat is applied to a solid, causing its molecules to move quickly.

5.A) strontium, sodium, copper, potassium

6 0
2 years ago
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