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Lena
3 months ago
9

Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el

ectric field at the center of the square? (k=1/4πϵ0=8.99×109N⋅m2/C2 )
Express your answer in terms of the variables d, q, and appropriate constants.
Physics
2 answers:
Ostrovityanka [3.2K]3 months ago
8 0
<span>this might be useful
Regarding the field, the two charges placed opposite cancel each other out!
Therefore, E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
given k = 8.99×10^9 N·m²/C²,
E = 1.789×10¹⁰ N·m²/C² * Q / d² </span>
Ostrovityanka [3.2K]3 months ago
7 0

Answer:

E = 2kq / d²

Explanation:

Let the side length of the square be d.

The distance from the center to a corner, a, is calculated as:

a = √(d² + d²) / 2

a = √(2d²) / 2

a = (√2 * d) / 2

a = d / √2

Electric field magnitude is:

E = kq / a²

Substituting a = d / √2, yields:

E = kq / (d / √2)²

E = kq / (d² / 2)

E = 2kq / d²

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I do not agree. Many materials may fluoresce when exposed to ULTRAVIOLET light, not in microwaves.:)
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An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t
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Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension present in rope = 550 N

The total torques will be

-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m

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3 months ago
The height of a typical playground slide is about 6 ft and it rises at an angle of 30 ∘ above the horizontal.
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Answer:

The coefficient of kinetic friction is found to be 0.432.

Explanation:

Comprehensive steps and derivations with necessary substitutions are detailed in the attached document.

6 0
3 months ago
Albert presses a book against a wall with his hand. As Albert gets tired, he exerts less force, but the book remains in the same
Maru [3345]

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.

7 0
3 months ago
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