When a positively charged object is brought close to 2 metal spheres that are touching, what will happen to the electrons in the spheres?.

Let's assume the metal spheres are solid metal, and not charged beforehand.

The premise are similar, but the effect changes depending if you are approching from the side, between the spheres (left image) or the front (right image). Let's image the charged object is the tip of the arrow in the drawing, and it's small enough to be considered a point

In either case, the electron cloud in the spheres will be affected by the electric field generated by the charge, and gets attracted towards it (since the charge is positive and the electrons have by definition negative charge), moving according to the blue lines. For as long as the charge is in place, the "grouping" of charges on one side and the absence of charges on the other will create an electric dipole with the positive charge away from the side the arrived and the negative on the opposite end of the diameter.

The interesting part happens the moment you split the two spheres.

In the left case will realign based on the field lines (remember that a single charge will generate a radial field) and the moment you remove the charge the electrons are no longer drawn to one end, and will eventually re-distribute on the whole sphere, canceling the dipole.

In the right case instead, the fact that the spheres were touching allows the electrons in one of the sphere to migrate to the other, generating a net positive charge in the far sphere, and a net negative charge in the close one. if you separate the two spheres before removing the charge, the electrons are still "trapped" in the leftmost sphere, thus keeping it charged even after the original source is removed from the system.

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.

Answer:

Step-by-step explanation:

Time taken by the tomatoes to each the ground

using h = 1/2 g t^2 

t^2 = 2h/g = 2 x 50/ 9.8 = 10.2

t = 3.2 sec 

horizontal ditance = speed x time = 3 x 3.2 = 9.6 meters