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3 years ago

Explain how carbon is cycled between the hydrosphere and geosphere. Use specific examples.

1 answer:

5 0

Carbon is found in the solid form in geosphere of our earth. Coal and oil are some of the examples of materials containing carbon in the geosphere. when the coal or oil is burnt, carbon dioxide is formed and released in atmosphere. This carbon dioxide is absorbed by the water of the hydrosphere with the help of algae and plankton. The water turns acidic in nature. This way carbon is transferred from geosphere to hydrosphere.

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Which is not a property of a pure substance

Ann [662]

air is a mixture and not a pure substance.

5 0

3 years ago

A very thin oil film (n = 1.25) floats on water (n = 1.33). What is the thinnest film that produces a strong reflection for gree

mixer [17]

Answer:

200 nm is the thinnest film that produces a strong reflection for green light with a wavelength of 500 nm

Explanation:

If two reflected waves interfere constructively ,strong reflection is produced. Two reflected waves will experience a phase change

For constructive interference

$2\times n\times t=m\lambda$

for thinnest film m=1

refractive index should be taken for film n=1.25

thickness of the thinnest film is

$t=\frac{m\lambda}{2n} \\t=\frac{1\times 500}{2\times 1.25} \\t=200 nm$

6 0

3 years ago

Question 14

Nimfa-mama [501]

Answer:

The answer is Dependent Variable

5 0

3 years ago

11 kg is a familiar weight for a bag of flour. You are baking cookies for a Save The Rain Forest fund drive. It takes 500 g of f

arlik [135]

Answer: 22 batches.

Explanation:

Given that 11 kg is a familiar weight for a bag of flour. Also, it is given that It takes 500 g of flour to make one batch of cookies.

How many batches of cookies can you make with one bag of flour

Let's first convert 11 kg into grams (g) by multiplying it by 1000

11 × 1000 = 11000 g

Divide 11000 by 500

11000/500 = 22

Therefore, 22 batches of cookies can be made with one bag of flour.

8 0

3 years ago

A person is standing on an elevator initially at rest at the first floor of a high building. The elevator then begins to ascend

GREYUIT [131]

Answer:

The found acceleration in terms of h and t is:

$a=\frac{h}{5(t_1)^2}$

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

Constant acceleration, starts from rest.

Distance = $y = \frac{1}{2}a(t_1)^2$

Velocity = $v_1=at_1$

<h3>Stage 2</h3>

Constant velocity where

Velocity = $v_o=v_1=at_1$

Distance =

<h3> $y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\$ </h3><h3 /><h3>Stage 3</h3>

Constant deceleration where

Velocity = $v_0=v_1=at_1$

Distance =

$y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2$

<h3>Total Height</h3>

Total height = y₁ + y₂ + y₃

Total height = $\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2$

<h3 /><h3>Acceleration</h3>

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

$a=\frac{h}{5(t_1)^2}$

6 0

3 years ago