Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

The answer is letter d, 8.3.

Here’s a solution for the given problem:

We have:

B = 6i - 8j 

Let A be unknown; we'll denote A as = mi + nj 

The resultant A+B lies along the x-axis (which implies A+B = Ki + 0j, where K is yet to be determined,

and we also know the magnitude of A+B is equivalent to the magnitude of A,

therefore, mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2. 

Using vector addition, A+B becomes (m+6)i + (n-8)j.

Since we know A+B = Ki + 0j, we can establish that: 

m + 6 = K 

n - 8 = 0, which gives n=8. 

Thus, K^2=m^2+n^2 means (m+6)^2 = m^2 +8^2 

= m^2 + 12m + 36 = m^2 + 64 

which gives us 12m = 28 

m = 2.33333... 

Consequently, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333.

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.