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Mamont248 [21]
3 years ago
13

What is the net force on charge placed at the center of square of length 1m? All charges have same magnitude of 2 micro coulomb.

Physics
1 answer:
Masja [62]3 years ago
5 0

Answer:

Electric Field intensity is zero.

The reason for that is:

All charges are placed at equal distances from the center of the square and have same magnitude and sign. This means they will exert equal and opposite forces on the test charge at the center. Net force will become Zero.

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Which method will correctly determine whether the forces on an object are balanced or unbalanced?
MaRussiya [10]

For this case, the first thing you should do is define a reference system.

Once the system is defined, we must follow the following steps:

1) Do the sum of forces in a horizontal direction

2) Do the sum of forces in vertical direction

The forces will be balanced if for each direction the net force is equal to zero.

The forces will be unbalanced if for each direction the net force is nonzero.

Answer:

Add the forces in the horizontal and vertical directions separately.

6 0
3 years ago
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Do any of the atom diagrams below represent atoms of the same element?
sergeinik [125]

Atom A and atom C are the same element.

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2 years ago
a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec
MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.  

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

\frac{1}{2}  mu_1^2+0=\frac{1}{2}  mv_1^2+\frac{1}{2}  mv_2^2

∴  \frac{1}{2}  m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2

∴ \frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2}  mv_1^2+\frac{1}{2} mv_2^2

∴ mv₁v₂ = 0

  1. It is impossible for the mass to be zero.
  2. Because the second ball moves, velocity v2 cannot be zero.
  3. As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

#SPJ4

3 0
1 year ago
What is the kinetic energy of a baseball with a mass of 0.133 kg and moving at a speed of 28.7m/s
blsea [12.9K]

Answer:

54.7 J

Explanation:

kinetic energy formula

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3 years ago
the magnitude of the normal force acting on a person with mass of 70 kg standing at rest on the flat ground would be ?
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Answer:

f = mg \\  = 70 \times 9.8 =  |f|

3 0
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