The answer would be Exosphere because, there are 3 main regions that circulate oxygen through the Earths system, which are the Biosphere, Atmosphere, and the Lithosphere.

5 0

2 years ago

With detailed explaniation

belka [17]

Ø=37°
Initial velocity=u=20m/s
g=10m/s²

$\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}$

$\\ \rm\Rrightarrow H_{max}=20sin^237$

$\\ \rm\Rrightarrow H_{max}=7.2m$

$\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}$

$\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}$

$\\ \rm\Rrightarrow R=40sin74$

$\\ \rm\Rrightarrow R=38.5m$

$\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}$

$\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}$

$\\ \rm\Rrightarrow T=4sin37$

$\\ \rm\Rrightarrow T=2.4s$

Now

$\\ \rm\Rrightarrow v=u-gt$

$\\ \rm\Rrightarrow v=20-10(2.4)$

$\\ \rm\Rrightarrow v=20-24$

$\\ \rm\Rrightarrow v=-4m/s$

4 0

2 years ago

If two particles have equal charges and are placed near one another, how

navik [9.2K]

Answer: C) The two particles will move away from each other

Explanation:

When two electrically charged bodies come closer, appears a force that attracts or repels them, depending on the sign of the charges of this two bodies.

This is stated by Coulomb's Law:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

Mathematically this law is written as:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$

Where $K$ is a proportionality constant.

Now, if $q_{1}$ and $q_{2}$ have the same sign charge (both positive or both negative), a repulsive force will act on these charges.

5 0

3 years ago