All categories

3 years ago

What is Earth's role in the hierarchy of organization within the universe

1 answer:

7 0

In the hierarchy of Organization within the universe , currently Earth placed as the 3rd Planet (after Mercury and Venus) , and the only planet that provided living organisms with enough resources to live such as water, foods, breathing air, etc.

You might be interested in

When a bond Norma between two or more elements, what part of the atom is forming the bond?

Mumz [18]

The electrons are the only part of the atom that forms the bond

7 0

4 years ago

PLEASE HELP WILL GIVE BRANLIEST IF CORRECT

coldgirl [10]

Answer:

Explanation:

z: commutator
y: brush
x: armature

3 0

3 years ago

Read 2 more answers

In atoms, electrons surround the nucleus in

nasty-shy [4]

In atoms, electrons surround the nucleus in specific energy levels.

Answer: Option B

<u>Explanation: </u>

Atom, actually considered as the tiny part that is ever present in this universe. Many theories and experiments were conducted, for studying what was present inside of an atom, and many theories came into light.

And finally, Bohr-Sommerfeld Theory stated the final conclusion, that the positive charge is at the centre of the nucleus, and all the electrons revolve in their specific energy levels. If an atom wants to move to lower energy state, it should emit energy, whereas for going to higher state, it should gain energy.

8 0

3 years ago

Determine Force. 589J of work was accomplished over a distance of 1100m, how much force was exerted? Include the correct units.

LiRa [457]

Answer:

Explanation:

Given:

A = 589 J

D = 1 100 m

____________

F - ?

A = F·D

F = A / D = 589 / 1100 ≈ 0.54 N

5 0

1 year ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago