Solid mixtures if it has particles that are _______ _____

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

1. Which of the following is not an acceptable Physical Education activity?

Dmitrij [34]

The answer is A

Explanation:
Vacuuming doesn’t involve a lot of physical movements.

3 0

2 years ago

Read 2 more answers

“I know what the atomic number of this atom is, but I don’t know what the number of electrons is,” a friend says. How would you

liq [111]

Once the atomic number of an atom is known, the number of electrons can be deduced depending on if the atom is an ion or a neutral one.

<h3>Atomic number</h3>

The atomic number of an atom is the number of protons in the nucleus of the atom.

For atoms that are neutral, that is, no net charges, the number of protons is always equal to the number of electrons. In other words, the positive charges always balance the negative charges in neutral atoms.

Thus, if the atomic number of a neutral atom is 6, for example, the proton number will also be 6. Since the proton must balance the electron, the number of electrons will also be 6.

More on atomic numbers can be found here; brainly.com/question/17274608

8 0

1 year ago

Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?

lesantik [10]

The equilibrium conditions allow to find the results for the balance forces are:

F₁ = 225.4 N

F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

∑ F = 0

∑ τ = 0

Where F are the forces and τ the torques.

The torque is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

F₂ 2 - W₁ 1 - W₂ 1.5 = 0 $\frac{W_1 \ 1 + W_2 \ 1.5}{2}$