1.In a pith ball experiment, the two pith balls are at rest. The magnitude of the tension in each string is |T| = 0.55 N, and the angle between each string and a vertical line is θ = 27.33°. What are the values for the magnitudes of electrostatic force, Fq, and the gravitational force, Fg?

2.One object has a charge of +5.0 × 10−6 C, and a second object has a charge of +2.0 × 10−6 C. The objects are 0.5 meters apart. What is the electrostatic force between them?

Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.

For first question make a FBD diagram and then solve components.

For the second question remember the basics F= kq1q2/(r)^2

1. To find the magnitudes of the electrostatic force, Fq, and the gravitational force, Fg, we can use the given information.

First, let's find the gravitational force, Fg. The formula for gravitational force is given by:

Fg = m * g

where m is the mass of the object and g is the acceleration due to gravity.

In this case, the pith balls are at rest, so we can assume that the net force acting on them is zero. Therefore, the gravitational force pulling the pith balls downward is balanced by the tension in the strings.

Since the tension in each string is |T| = 0.55 N, and the angle between each string and the vertical line is θ = 27.33°, we can use trigonometry to find the vertical component of the tension:

T_vertical = |T| * cos(θ)

T_vertical = 0.55 N * cos(27.33°)

T_vertical = 0.55 N * 0.891 = 0.49 N (approximately)

The gravitational force acting on each pith ball is equal to the tension's vertical component, so we have:

Fg = 0.49 N (approximately)

Next, let's find the electrostatic force, Fq. The formula for electrostatic force is given by:

Fq = k * |q1| * |q2| / r^2

where k is the electrostatic constant (k = 9.0 x 10^9 N m^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between the charges.

Given that one object has a charge of +5.0 × 10−6 C and the other object has a charge of +2.0 × 10−6 C, and the distance between them is 0.5 meters, we can substitute these values into the formula:

Fq = (9.0 x 10^9 N m^2/C^2) * (5.0 × 10−6 C) * (2.0 × 10−6 C) / (0.5 m)^2

Fq = (9.0 x 10^9 N m^2/C^2) * (10 × 10^-12 C^2) / (0.25 m^2)

Fq = (9.0 x 10^9 N m^2/C^2) * 40 × 10^-12 C^2 / 0.25 m^2

Fq = (9.0 x 10^9 N m^2/C^2) * 1600 × 10^-12 C^2 / 0.25 m^2

Fq = (9.0 x 10^9 N m^2/C^2) * 6400 × 10^-12 C^2 / 0.25 m^2

Fq = 2304000 × 10^-3 N (approximately)

Fq = 2304 N (approximately)

Therefore, the values for the magnitudes of the electrostatic force Fq and the gravitational force Fg are:

Fq ≈ 2304 N
Fg ≈ 0.49 N

2. To calculate the electrostatic force between two charged objects, we can use Coulomb's Law:

Fq = k * |q1| * |q2| / r^2

where k is the electrostatic constant (k = 9.0 x 10^9 N m^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between the charges.

Given that one object has a charge of +5.0 × 10−6 C and the other object has a charge of +2.0 × 10−6 C, and the distance between them is 0.5 meters, we can substitute these values into the formula:

Fq = (9.0 x 10^9 N m^2/C^2) * (5.0 × 10−6 C) * (2.0 × 10−6 C) / (0.5 m)^2

Fq = (9.0 x 10^9 N m^2/C^2) * (10 × 10^-12 C^2) / (0.25 m^2)

Fq = (9.0 x 10^9 N m^2/C^2) * 40 × 10^-12 C^2 / 0.25 m^2

Fq = (9.0 x 10^9 N m^2/C^2) * 1600 × 10^-12 C^2 / 0.25 m^2

Fq = (9.0 x 10^9 N m^2/C^2) * 6400 × 10^-12 C^2 / 0.25 m^2

Fq ≈ 2304000 × 10^-3 N

Fq ≈ 2304 N

Therefore, the electrostatic force between the two charged objects is approximately 2304 N.

To find the values for the magnitudes of electrostatic force (Fq) and gravitational force (Fg) in the pith ball experiment, we can use the following formulas:

1. For electrostatic force (Fq): Fq = k * (|q1| * |q2|) / r^2
2. For gravitational force (Fg): Fg = G * (|m1| * |m2|) / r^2

Where:
- k is the electrostatic constant (8.99 × 10^9 N * m^2 / C^2)
- q1 and q2 are the charges of the objects in Coulombs (C)
- r is the distance between the objects in meters (m)
- G is the gravitational constant (6.67430 × 10^-11 N * m^2 / kg^2)
- m1 and m2 are the masses of the objects in kilograms (kg)

Let's solve these questions one by one:

1. In the pith ball experiment:
- We are given |T| = 0.55 N (tension in each string) and θ = 27.33° (angle between each string and a vertical line).
- The tension in the strings can be considered as the electrostatic force acting on the pith balls.
- The electrostatic force is balanced by the gravitational force acting on the pith balls.

Using the given information, we can calculate the magnitudes of the electrostatic force (Fq) and gravitational force (Fg) as follows:

- Fq = |T| = 0.55 N (Since the electrostatic force is equal to the tension in each string)
- Fg = |m| * g (Where m is the mass of the pith ball and g is the acceleration due to gravity)

2. For the electrostatic force between the two charged objects:
- We are given the charges of the two objects (q1 = +5.0 × 10^−6 C and q2 = +2.0 × 10^−6 C) and the distance between them (r = 0.5 m).

Using the formula mentioned earlier, we can calculate the electrostatic force (Fq) as follows:

- Fq = k * (|q1| * |q2|) / r^2

Substituting the given values into the formula, we can find the answer.

Note: The mass of the pith ball and the gravitational force may not be directly related to the electrostatic force between the charged objects. However, they are mentioned to explain the pith ball experiment and highlight the difference in calculations for electrostatic and gravitational forces.