How many electrons must be removed from a neutral, isolated conducting sphere to give it a positive charge of 8.0 x 10 8 C? [Q=n

In the query, it is stated that

     The distance the electron covers is  

Typically, the force exerted on this electron is expressed mathematically as

Where F signifies the force and  q represents the charge of the electron, which is a fixed value of

    Thus  

Generally, the work-energy theorem is mathematically framed as

Where W denotes the work done on the electron by the electric field and    is the change in kinetic energy

Additionally, work done on the electron can also be described as

Where   assuming that the electron's movement aligns with the x-axis  

        So

Inserting values

According to the conservation of energy

Where signifies the change  in  potential energy  

Thus  

Answer:

(A) The tension's magnitude grows to four times the initial value, 4F.

Explanation:

When an object travels in a circular path, a centripetal force is exerted upon it. In this instance, the centripetal force acting on the stone can be represented by .

                   Here, m denotes the mass of the object

                               v is the velocity or speed of the object

                               r signifies the radius of the circular path

Importantly, the tension corresponds to the centripetal force.

Initially, the string completes one revolution each second, and subsequently, it accelerates to perform two revolutions in the same time frame. This signifies that the speed has increased twofold.

Applying our formula:F =

                               where F indicates the tension in the string

assuming the starting speed is v, after doubling it becomes 2v

Maintaining the circle's radius, we arrive at:

F=

From this equation, it's clear that the initial tension has quadrupled.

Consequently, the magnitude of the tension increases to four times its original value, 4F.

Solution:

/ Em₀ = 0.30

Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

      K = ½ m v²

      U = mgh

We assess the mechanical energy at two positions:

Initial. Lower

    Em₀ = K = ½ m v²

At its highest point

    = U = mg and

Now let's compute

    Em₀ = ½ m 3.6²

    Em₀ = m 6.48

    = m 9.8 × 0.2

    = m 1.96

Thus the energy lost is given by:

    / Em₀ = m 1.96 / m 6.48

    / Em₀ = 0.30

This means that 30% of the sun's energy is transformed into potential energy.

There are various conversion possibilities.

This energy changes into thermal energy affecting the spores and air, since it cannot be regained.