Imagine a universe in which, like in ours, there are two kinds of charges (positive and negative), with the like charges repelling and unlike charges attracting each other, but with the electric force somewhat different from Coulomb's Law. Instead of varying with inverse distance squared, the force between charges in this imagined world would vary with the inverse distance cubed. In such a universe, would the static charges still always be distributed on the surfaces of conductors?

the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude

Explanation:

In this hypothetical system the electric force is of type

       F =

in this case the force decays to zero much faster,

if we call Fo the force of Coulomb's law

         F₀ =

assuming the constant k is the same

the relationship between the two forces is

        F / F₀ = 1 / r

        F = F₀ / r

when analyzing this expression the force decays much faster to zero.

In an electric conductor, charges of the same sign may not feel any repulsive force from other charges that are at a medium distance, so there is a probability that some charges are distributed in the volume of the material, this does not happen with coulomb's law

Consequently, the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude

Independent variable: the best method to get rid of them.

Dependent variable: washing with soap and water.

Hypothesis: Organic oils

Control group: water

Answer:

Step-by-step explanation:

D.
Heat each cube to the same temperature, place each cube into different containers with 500 grams of water at the same temperature, and measure the temperature of the water.

Answer:

Step-by-step explanation:

To solve this problem, we can use the conservation of energy and conservation of momentum principles.

Conservation of energy:

The total initial energy is the rest energy of the proton and neutron, which is given by:

Ei = (mp + mn)c^2

where mp and mn are the masses of the proton and neutron, respectively, and c is the speed of light.

The total final energy is the rest energy of the deuteron plus the energy of the gamma ray, which is given by:

Ef = (md)c^2 + Eg

where md is the mass of the deuteron and Eg is the energy of the gamma ray.

According to the conservation of energy principle, the initial energy and final energy must be equal, so we have:

Ei = Ef

(mp + mn)c^2 = (md)c^2 + Eg

Conservation of momentum:

The total initial momentum is zero because the proton and neutron are at rest. The total final momentum is the momentum of the deuteron and the momentum of the gamma ray. Since the gamma ray is massless, its momentum is given by:

pg = Eg/c

where pg is the momentum of the gamma ray.

According to the conservation of momentum principle, the total final momentum must be equal to zero, so we have:

0 = pd + pg

where pd is the momentum of the deuteron.

Solving for md and pd:

From the conservation of energy equation, we can solve for md:

md = (mp + mn - Eg/c^2)/c^2

Substituting this expression into the conservation of momentum equation, we get:

pd = -pg = -Eg/c

Substituting the given values, we have:

mp = 1.6726 × 10^-27 kg mn = 1.6749 × 10^-27 kg Eg = 2.2 × 10^6 eV = 3.52 × 10^-13 J

Using c = 2.998 × 10^8 m/s, we get:

md = (1.6726 × 10^-27 kg + 1.6749 × 10^-27 kg - 3.52 × 10^-13 J/(2.998 × 10^8 m/s)^2)/(2.998 × 10^8 m/s)^2 = 3.3435 × 10^-27 kg

pd = -Eg/c = -(3.52 × 10^-13 J)/(2.998 × 10^8 m/s) = -1.1723 × 10^-21 kg·m/s

Therefore, the mass of the deuteron is 3.3435 × 10^-27 kg, and its momentum is -1.1723 × 10^-21 kg·m/s.

The question specifies the diameter of the screw, therefore the IMA of this screw is 0.812? / 0.318 = 8.02