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2 years ago

The carbon cycle relies on producers and consumers to survive. What is the role of each in the carbon cycle?

1 answer:

8 0

ANSWER: (ROLE of Producers in the cycle:totake in carbon dioxide from the atmosphere during photosynthesis and release carbon dioxide during respiration,ROLE of Consumers:to contribute to the carbon cycle when they perform respiration and expel carbon dioxide into the atmosphere)
Hope this Helps SOMEONE!!
,Thank you

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Two rocks are moving through empty space in a straight line at the same speed. One has a mass of 10 kg, the other has a mass of

Yuki888 [10]

A and D are definitely wrong because the two rocks have different masses, so it leaves us with B and C. The most logical answer to the question is
C <span>The one with greater mass takes more force to stop.
</span>

6 0

3 years ago

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

Distinguish among the three places that volcanoes occur with respect to plate boundaries

Olin [163]

Divergent
Convergent
And

5 0

3 years ago

An object is placed at 0 on a number line. It moves 3 units to the right, then 4 units to the left, and then 6 units to the righ

Verdich [7]

The correct answer is: +5

Explanation:

An object is placed at 0; it means:

Initial position of the object = 0.

Now it moves to 3 units to right, so keeping the standard cartesian coordinate system in mind (in right right x-axis is positive and left x-axis is right), the new position of the object will be +3.

Object now moves 4 units to the left, it means +3 - 4 = -1; object is at the position -1.

Object then moves 6 units to the right, therefore,

Final position of the object = -1 + 6 = +5.

Displacement = Final position - Initial position

Displacement = +5 - 0 = +5

6 0

3 years ago

Read 2 more answers

I only need help on #15! Thanks!

Vanyuwa [196]

I would have to say that 'B' and 'D' are both correct.

Increasing the voltage that you're using to operate a circuit
causes the current in the circuit to increase. But current is
just the number of electrons that are flowing through it. So
right there, you have the increase in the number of charges.

Now, every electron that flows through the circuit gives up
some energy on the way. So if there are more electrons
making the trip, then more energy has been put into the circuit.

Jessica, I absolutely love your printing.
I wish I could print so clearly.

7 0

3 years ago