Answer:Thus, if each of the charges were reduced by one-half, the repulsion would be reduced to one-quarter of its former value. Also In electrostatics, the electrical force between two charged objects is inversely related to the distance of separation between the two objects. ... And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects.
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Explanation:
Nothing, until those two objects physically touch each other; contact aligns polarity among the now single shared mass.
(Your question never states if both objects are unique or similar polar charges, so I just assumed they were both neutral objects existing within an electric field.)
So a better question would then be, what is gravity’s relationship with an electric field?
You could solve this with the following: confine the electric field’s volume to a set variable (never increasing nor decreasing in size or shape); density is variable and easily definable. This creates the limit to build upon. This density has to be fluid and has electron mass (full of electrons at rest mass, so with substance but no movement). Within, create a closed system (the hard part in this equation; outside interference like ambient light will eschew results) where each variable of kinetic energy then is accounted for or measurable (including heat and light, and the physical movement of the two objects)
Determine the mass for both objects, calculate gravity for both and each body’s inertia on the other as a sum over distance. record results. Polarity is shared across the masses until there is no longer inertia (one mass or contact).
Explanation:
It is given that,
Mass, 
Acceleration due to gravity, 
Acceleration due to gravity is given by :
r = distance from head
r = 1.42 m
So, at the distance of 1.42 meters from the head its gravitational pull on you match that of Earth's. Hence, this is the required solution.
The electric potential at the origin of the xy coordinate system is negative infinity
<h3>What is the electric field due to the 4.0 μC charge?</h3>
The electric field due to the 4.0 μC charge is E = kq/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q = 4.0 μC = 4.0 × 10 C and
- r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m
<h3>What is the electric field due to the -4.0 μC charge?</h3>
The electric field due to the -4.0 μC charge is E = kq'/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q' = -4.0 μC = -4.0 × 10 C and
- r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m
Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is
E" = E + E'
= -2E
= -2kq/r²
<h3>What is the electric potential at the origin?</h3>
So, the electric potential at the origin is V = -∫₂⁰E".dr
= -∫₂⁰-2kq/r².dr
Since E and dr = dx are parallel and r = x, we have
= -∫₂⁰-2kqdxcos0/x²
= 2kq∫₂⁰dx/x²
= 2kq[-1/x]₂⁰
= -2kq[1/x]₂⁰
= -2kq[1/0 - 1/2]
= -2kq[∞ - 1/2]
= -2kq[∞]
= -∞
So, the electric potential at the origin of the xy coordinate system is negative infinity
Learn more about electric potential here:
brainly.com/question/26978411
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Answer:
length of the ladder is 13.47 feet
base of wall to latter distance 6.10 feet
angle between ladder and the wall is 26.95°
Explanation:
given data
height h = 12 feet
angle 63°
to find out
length of the ladder ( L) and length of wall to ladder ( A) and angle between ladder and the wall
solution
we consider here angle between base of wall and floor is right angle
we apply here trigonometry rule that is
sin63 = h/L
put here value
L = 12 / sin63
L = 13.47
so length of the ladder is 13.47 feet
and
we can say
tan 63 = h / A
put here value
A = 12 / tan63
A = 6.10
so base of wall to latter distance 6.10 feet
and
we say here
tanθ = 6.10 / 12
θ = 26.95°
so angle between ladder and the wall is 26.95°
Answer:
Explanation:
The rotation rate of the man is:
The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:
![(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega](https://tex.z-dn.net/?f=%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%5Ccdot%20%280.16%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%29%20%3D%20%5B%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%2B20000%5C%2Ckg%5Ccdot%20m%5E%7B2%7D%5D%5Ccdot%20%5Comega)
The final angular speed is:
