500mL of He at 98 kPa expands to 750 mL. Find P2

Response:

E =  ρ ( R1²) / 2 ∈o R

Clarification:

Provided information

Two cylinders are aligned parallel

Distance = d

Radial distance = R

d < (R2−R1)

To determine

Express the response using the variables ρE, R1, R2, R3, d, R, and constants

Solution

We have two parallel cylinders

therefore, area equals 2 R × l

And we apply Gauss's Law

EA = Q(enclosed) / ∈o......1

Initially, we calculate Q(enclosed) = ρ Volume

Q(enclosed) = ρ ( R1² × l )

Thus, inserting all values into equation 1

produces

EA = Q(enclosed) / ∈o

E(2 R × l)  = ρ ( R1² × l ) / ∈o

This simplifies to

E =  ρ ( R1²) / 2 ∈o R

Answer:

Explanation:

The data indicates that point A is located midway between two charges.

To calculate the electric field at point A, we begin with the field produced by charge -Q ( 6e⁻ ) at A:

= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴

= 13.82 x 10⁻⁶ N/C

This field points towards Q⁻.

A similar field will arise from the charge Q⁺, but it will direct away from Q⁺ toward Q⁻.

To find the resultant field, we add these contributions:

= 2 x 13.82 x 10⁻⁶

= 27.64 x 10⁻⁶ N/C

For the force acting on an electron placed at A:

= charge x field

= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶

= 44.22 x 10⁻²⁵ N

Answer:

the maximum static friction force of the wall acting on the book (Increasing)

the normal force of the wall acting on the book (Decreasing)

the weight of the book (Constant)

Explanation:

According to Newton's third law of motion:

"Every action has an equal and opposite reaction"

In the scenario provided, Albert is pressing the book against the wall and subsequently decreases the force applied against the wall.

Let's evaluate all forces influencing the book in this situation.

1. Weight of the book acting downwards (y-axis)

2. Friction from the book against the wall acting upwards (y-axis)

3. Albert’s force exerted on the book against the wall (x-axis)

4. Normal force of the wall reacting to Albert’s applied force (x-axis)

As Albert eases off his force, the new scenario reads:

1. The weight remains constant as represented by W = mg

Since neither mass nor gravitational acceleration has changed, the weight exerted on the book remains the same.

2. As Albert reduces his force, the wall’s normal reaction force decreases correspondingly, following Newton's third law of motion.

3. Friction operates in response to the force applied to it. With a box resting on the floor, no friction acts upon it until it is dragged, at which point friction begins to manifest and rise until it reaches its maximum. Therefore, when Albert diminishes his force, the weight's pull will influence the book and the maximum static friction will rise to resist the book’s downward movement.

It should be noted that the maximum static friction is working to prevent movement of the book. With Albert's force reduced, but the weight of the book unchanged, maximum static friction increases to prevent downward movement.

Response:

n (a sin θ) = m λ₀

Since n > 1, this indicates that the fringes separate further apart

Clarification:

In a diffraction experiment, the equation for constructive interference fringes is provided by

a sin θ = m λ₀

It is presumed that the air has been evacuated from the experiment, setting n = 1

When this experiment is conducted in water, the wavelength alters

λₙ = λ₀ / n

for achieving constructive interference

a sin θ = m λₙ

we replace

a sin θ = m λ / n

n (a sin θ) = m λ₀

Given that water's refractive index is n = 1.33, the distance between the fringes increases due to n > 1, causing the fringes to move apart