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IRINA_888 [86]
3 years ago
13

Why is electrostatic force able to act at a distance ​

Physics
1 answer:
mote1985 [20]3 years ago
7 0
In Electrostatics the electrical force between Two charged objects is inversely Related to the distance of separation between the two objects .
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Papessa [141]
Vi= square root vi^2 -2ad
7 0
3 years ago
A student attaches a rope to a 20.0 kg box of books. He pulls with a force of 90.0 N at an angle of 30 degrees with the horizont
kolbaska11 [484]

Answer:

The magnitude of the acceleration of the box is 2.01 m/s².

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following horizontal forces:

Fr = friction force.

Fx = Horizontal component of the applied force, F.

And we have the following vertical forces:

Fy = vertical component of the applied force.

N = normal force exerted on the box.

W = weight of the box.

According to Newton´s second law:

∑F = m · a

Then, in the horizontal direction:

Fx - Fr = m · a

Where "m" is the mass of the box and "a" its acceleration.

Fx can be obtained by trigonometry (see figure):

Fx = F · cos 30°

Fx = 90.0 N · cos 30°

Fr is calculated as follows:

Fr = μ · N

Where μ is the coefficient of friction and N the normal force.

So, we have to find the magnitude of the normal force.

Using Newton´s second law in the vertical direction:

∑F = N + Fy - W = m · a

Notice that the box has no vertical acceleration, then:

N + Fy - W = 0

Solving for N:

N = W - Fy

The weight is calculated as follows:

W = m · g

Where g is the acceleration due to gravity:

W = 20.0 kg · 9.8 m/s² = 196 N

And the vertical component of the applied force can be obtained by trigonometry:

Fy = F · sin 30°

Fy = 90.0 N · sin 30°

The normal force will be:

N = W - Fy = 196 N - 90.0 N · sin 30°

N = 151 N

Now, we can calculate the friction force:

Fr = μ · N

Fr = 0.250 · 151 N

Fr = 37.8 N

And now, we can obtain the acceleration of the box:

Fx - Fr = m · a

(Fx - Fr) / m = a

(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a

a = 2.01 m/s²

The magnitude of the acceleration of the box is 2.01 m/s².

8 0
3 years ago
A 20 kg object is floating in space. What is its mass?
trapecia [35]

It's still 20 kg. Mass doesn't change according to gravity, only weight changes.

5 0
3 years ago
At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at
Agata [3.3K]
We have the following equation for height:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 Where,
 a: acceleration
 vo: initial speed
 h0: initial height.
 The value of the acceleration is:
 a = -g = -9.8 m / s ^ 2
 For t = 0 we have:
 h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
 h (0) = h0
 h0 = 0 (reference system equal to zero when the ball is hit).
 For t = 5.8 we have:
 h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
 (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
 vo = (1/2) * (9.8) * (5.8)
 vo = 28.42
 Substituting values we have:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
 Rewriting:
 h (t) = -4.9 * t ^ 2 + 28.42 * t
 The maximum height occurs when:
 h '(t) = -9.8 * t + 28.42
 -9.8 * t + 28.42 = 0
 t = 28.42 / 9.8
 t = 2.9 seconds.
 Answer:
 
The ball was at maximum elevation when:
 
t = 2.9 seconds.
8 0
3 years ago
Read 2 more answers
An element's atomic number is the​
Zepler [3.9K]

the atomic number of a chemical element (also known as its proton number) is the number of protons found in the nucleus of an atom of that element, and therefore identical to the charge number of the nucleus.

Hope this helped

8 0
3 years ago
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