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Artemon [7]
3 years ago
9

I need to know everything about mars. I need to do a full 3 page essay. please help!

Physics
1 answer:
zheka24 [161]3 years ago
4 0

Answer:

Mars was the Roman god of War along with an agricultural guardian. He is most closely related to the god Ares of Greek Mythology. In Roman mythology, he was second in importance to Jupiter, Rome's god of the Skies and Weather. Jupiter was the king of the Roman Pantheon, husband of the queen of gods Juno. He was also Mars' father. Unlike his Greek Counterpart, Ares who was most known for his hot headed temper and associated with hate and anger, Mars was part of the Romans <em>Archaic Triad</em>, sort of like the Big Three of Greek religion. The members of said Triad included Mars, Jupiter, and Quirinus, who had no Greek equivalent. Mars was most commonly depicted in posed of valor and strength, carrying swords or shields. He wore common Roman armor, including the plumed helmet. He was pictured as a strong leader of the Roman Army. The fourth planet from the Sun was given the name Mars when it was first discovered because it was red, much like the main color the Roman god was affiliated with.

This was mostly just random facts but i hope it helped some with your essay :)

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Explanation:

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When preparing to strip insulation in order to splice conductors, a person should strip the insulation back far enough so that t
Fiesta28 [93]

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Closely fits into the connector.

Explanation:

It's one of the steps used for the splicing of aluminium conductors in the underground connections. Where we do the strip insulation to splice the conductors by using compression type connectors.

3 0
3 years ago
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
storchak [24]

Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

The formula for the force exerted between two charges is

F=k \dfrac{ q_1q_2}{r^2}

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

F=k\dfrac{q^{2}}{r^2}

\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

7 0
3 years ago
What displacement in cm would occur with a 75 N/m spring if you placed a 300 N weight on the spring?
ollegr [7]

Surface tension=75N/m

Weight=300N

\\ \bull\tt\longmapsto Surface\:Tension=\dfrac{Weight}{Displacement}

\\ \bull\tt\longmapsto Displacement=\dfrac{Weight}{Surface\:Tension}

\\ \bull\tt\longmapsto Displacement=\dfrac{300}{75}

\\ \bull\tt\longmapsto Displacement=4m

\\ \bull\tt\longmapsto Displacement=400cm

6 0
2 years ago
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240
Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

As

V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

3 0
3 years ago
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