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 :)
The resistance needed to be added is R
The Current is 2 ma
The voltage reading is a maximum of 50 volts.
The ma meter has an internal resistance of 40 ohms.
Formula
E = I * R
Givens
E = 50
I = 2 ms
R = R + 40
Solution
E = I * R
I = 2 ma [ 1 amp / 1000 ma] = 0.002 amp
50 = 0.002 * (R + 40) Divide by 0.002
50/0.002 = R + 40
25000 = R + 40 Subtract 40 from both sides.
R = 25000 - 40
R = 24960 Answer
Answer is miles sir your welcome it was simple
Answer:
The answer is A and D but if you want one anwer i think the anwer is A
Explanation:
Answer:
0.08 N/C
Explanation:
Electric Field: This is defined as the force per unit charge exerted at a point. The expression for electric field is given as,
E = Kq/r².............................. Equation 1
Where E = Electric Field, q = Charge, k = proportionality constant, r = distance.
making q the subject of the equation,
q = Er²/k............................... Equation 2
Given: E = 2 N/C, r = 4 m,
Substitute into equation 2
q = 2(4)²/k
q = 32/k C.
When r is increased to 20 m,
E = k(32/k)/20²
E = 32/400
E = 0.08 N/C.
Hence the electric Field = 0.08 N/C
