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givi [52]
3 years ago
11

How is it possible for one electrically neutral atom to exert an electrostatic force on other electrically neutral atom?

Physics
1 answer:
Free_Kalibri [48]3 years ago
8 0

Answer:

Explained

Explanation:

Although Atom are electrically neutral. But atom atom is combination of nucleus and electrons. The nucleus of the atom is composed neutron and positively charged protons. On the outside of nucleus at some distance are the electrons which are negatively charged. So, there is difference in position of the two differently charge species. So, this way a electrically neutral atom can exert a electrostatic force on other electrically neutral atom

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A 4.0 kg ball is traveling at 3.0 m/s and strikes a wall. The ball bounces off the wall with a velocity of 4.0 m/s in the opposi
trasher [3.6K]

Answer:

280 N

Explanation:

Applying Newton's third second law of motion,

F = m(v-u)/t................... Equation 1

Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.

Note: Let the direction of the initial velocity of the ball be positive

Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s

Substitute into equation 1

F = 4(-4-3)/0.1

F = 4(-7)/0.1

F = -28/0.1

F = -280 N.

Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball

3 0
3 years ago
How can working together help scientist achieve their goals?
abruzzese [7]
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7 0
3 years ago
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Elan Coil [88]

Answer:D

Explanation:

6 0
1 year ago
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What items can be classified as matter?
natita [175]
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4 0
3 years ago
A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (
kvv77 [185]

Answer:

 v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

Lowest starting point.

          Em₀ = K = ½ m v²

final point. Higher

          Em_{f} = U = m g h

Let's use trigonometry to lock her up

          cos 60 = y / L

          y = L cos 60

Height is the initial length minus the length at the maximum angle

           h = L - L cos 60

           h = L (1- cos 60)

energy is conserved

         Em₀ = Em_{f}

          ½ m v² = mgL (1 - cos 60)

         v = 2g L (1- cos 60)

 

let's calculate

          v² = 2 9.8 3.0 (1- cos 60)

          v = 29.4 m / s

6 0
3 years ago
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