Answer:
A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in the following figure. The particle, confined to move along the x axis, is moved a small distance x along the axis ( where x << a) and released. Show that the particle oscillates in simple harmonic motion with a frequency given by,
We may balance the forces in order to calculate the tension.
We know that there are no horizontal forces being applied. Next, the vertical forces are:
Weight (W) which is acting downwards
Tension in each wire (T) which is acting upwards
Because the sign is stationary, the downward force must equal the upward force. So we may write:
Weight = 4 * tension
W = 4T
100 = 4T
T = 25 N
The tension in each wire is 25 Newtons
Answer:
θ = 41.8º
Explanation:
This is an internal total reflection exercise, the equation that describes this process is
sin θ = n₂ / n₁
where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂
θ = sin⁻¹ n₂ / n₁
let's calculate
θ = sin⁻¹ (1.00 / 1.50)
θ = 41.8º
The answer would be number four. I'm sorry if I am too late. Byes.....
<h3>
B. True</h3>
"This was the idea that non-living objects can give rise to living organisms."
