A car changes its velocity from 40.0 m/s to -5.0 m/s in 5.0 seconds what was the car average acceleration

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

ludmilkaskok [199]

A

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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5 0

3 years ago

Name the type of reproduction process as shown in Fig 1 and Fig 2. State one point of difference between the two

likoan [24]

Figure 1= binary fission in amoeba

figure 2= budding in yeast

difference

1.Parent divides to form two daughter cells and itself gets disappeared in binary fission but in budding , a bud gets matured and detaches from the parent

6 0

3 years ago

the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is

cluponka [151]

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

3 0

3 years ago

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

The light of that star actually gives off

ICE Princess25 [194]

-I believe the star gives off energy-, With most stars, like our sun, hydrogen is being converted into Helium, a process which gives off energy that heats the star.

7 0

3 years ago

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