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2 years ago

The phosphorus cycle is the biogeochemical cycle that describes the movement of phosphorus through the lithosphere,

Physics

1 answer:

ludmilkaskok [199]2 years ago

5 0

most common source of phosphates in the soil is the weathering of the mineral apatite.

Explanation:

Apatite is a group of phosphate minerals like hydroxylapatite, fluorapatite, and chlorapatite. When apatite rocks weather, they release the phosphate minerals mainly in the form of PO₄ ³⁻ . These minerals become dissolved in water (hydrosphere), where they are readily available to plants and other organisms in the biosphere. The phosphates are taken up and used in biosynthesis. When these organisms die and become buried with sediments, the phosphate gets back to the lithosphere as sedimentary rock.

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Explain the difference between the terms "rotation" and "revolution"

Novay_Z [31]

Answer:

Explanation:

"Rotation" refers to an object's spinning motion about its own axis. "Revolution" refers the object's orbital motion around another object. For example, Earth rotates on its own axis, producing the 24-hour day. Earth revolves about the Sun, producing the 365-day year.

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1 year ago

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If the mass of a material is 87 grams and the volume of the material is 14 cm3, what would the density of the material be?

tresset_1 [31]

Answer:

6.214g/cm³

Explanation:

The question is on density of a material

Density=mass/volume

Given, mass=87grams and volume= 14 cm³ density=?

Density=m/v 87/14 =6.214g/cm³

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2 years ago

What is the meaning of viscosity?

Vera_Pavlovna [14]

the state of being thick, sticky, and semifluid in consistency, due to internal friction.

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2 years ago

shusha [124]

Answer: The unit of impulse is applied to an object produces an equivalent vector change in its linear momentum, also in the same direction.

Explanation:

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2 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

1 year ago