Answer:
With changing speed and/or in a circle
Answer:
Myocardium. That is the type. (srry i was in a rush hope this helps)
Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
<h3>What is electric force?</h3>
Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.



Hence the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
To learn more about the electric force refer to the link;
brainly.com/question/1076352
<span>Since there is no friction, conservation of energy gives change in energy is zero
Change in energy = 0
Change in KE + Change in PE = 0
1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0
1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0
(vf^2 - vi^2) = 2 x g x (hi - hf)
Since it starts from rest vi = 0
Vf = squareroot of (2 x g x (hi - hf))
For h1, no hf
Vf = squareroot of (2 x g x (hi - hf))
Vf = squareroot of (2 x 9.81 x 30)
Vf = squareroot of 588.6
Vf = 24.26
For h2
Vf = squareroot of (2 x 9.81 x (30 – 12))
Vf = squareroot of (9.81 x 36)
Vf = squareroot of 353.16
Vf = 18.79
For h3
Vf = squareroot of (2 x 9.81 x (30 – 20))
Vf = squareroot of (20 x 9.81)
Vf = 18.79</span>
<h2>The distance between students is 2.46 m</h2>
Explanation:
The force of attraction due to Newton's gravitation law is
F = 
Here G is the gravitational constant
m₁ is the mass of one student
m₂ is the mass of second student .
and r is the distance between them
Thus r = 
If we substitute the values in the above equation
r = 
= 2.46 m