The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:
<h3>What are perfectly elastic collision?</h3>
Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.
In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.
Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.
Since some of the Kinetic energy is converted to potential energy of the body;
Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.
Learn more about elastic collisions at: brainly.com/question/7694106
<span>To do this question, we need to know that momentum is conserved, meaning the overall velocity of the two balls has to be the same before and after the collision. </span>
<span>After collision... </span>
<span>Ball 1: 4.33m/s *cos 30 = 3.75 m/s (x-component) </span>
<span>4.33m/s * sin 30 = 2.165 m/s ( y-component) </span>
<span>Ball 2 (struck ball): 5 m/s - 3.75m/s = 1.25 m/s (x-component) </span>
<span>-2.165 m/s (y-component) note: it has to be in the opposite direction to conserve momentum </span>
<span>tan-1(2.165/1.25) = 60 degrees </span>
<span>Struck ball's velocity = sqrt(1.25^2 + 2.165^2) = 2.5 m/s at 60 degree with respect to the original line of motion. </span>
<span>Hope you understand!</span>
The electric force between two charged particles can be increased by decreasing the distance between the two particles.
<h3>How to increase electric force between two charged particles.</h3>
The technique of decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. while
increasing the separation distance between objects decreases the force of attraction or repulsion between the objects.
Read more on Electric Force:
brainly.com/question/17692887
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Answer:
a. The thickness of the wire is 2.5 mm.
b. The wire is 0.25 cm thick.
Explanation:
Number of turns of the wire = 10
The length of total turns = 25 mm
a. The thickness of the wire can be determined by;
thickness of the wire = 
= 
= 2.5 mm
Therefore, the wire is 2.5 mm thick.
b. To determine the thickness of the wire in centimetre;
10 mm = 1 cm
So that,
2.5 mm = x
x = 
= 0.25 cm
The wire is 0.25 cm thick.
Answer:
The divergence on the sensor shows the magnitude of the charges
Explanation:
This will increase as length increases since it is said to be proportional to the length. note that test charge is always positive and charge on the grid is positive as indicated (1 nC)
