If electromagnetic radiation acted like particles in the double-slit experiment, we would observe one bright band would appear in the center of the screen.
<h3>Bahavior of particles in double-slit experiment</h3>
In a double-slit experiment, single particles, such as photons, pass one at a time through a screen containing two slits.
The photons behave like wave and the constructive interfernce of the waves of these photons will generate a high amplitude wave seen as a bright band in the center of the screen.
Thus, if electromagnetic radiation acted like particles in the double-slit experiment, we would observe one bright band would appear in the center of the screen.
Learn more about double slit experiment here: brainly.com/question/4449144
The stretch of the spring is
The constant of the spring is k=15 N/m, so we can find the force produced by the weight by using Hook's law:
Answer:
Explanation:
a = F/m = 7500/2000 = 3.75 m/s²
v² = u² + 2as
s = (v² - u²) / 2a
s = (0² - 45²) / (2(-3.75))
s = 270 m
Given:
(Initial velocity)u=20 m/s
At the maximum height the final velocity of the ball is 0.
Also since it is a free falling object the acceleration acting on the ball is due to gravity g.
Thus a=- 9.8 m/s^2
Now consider the equation
v^2-u^2= 2as
Where v is the final velocity which is measured in m/s
Where u is the initial velocity which is measured in m/s
a is the acceleration due to gravity measured in m/s^2
s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.
Substituting the given values in the above formula we get
0-(20x20)= 2 x- 9.8 x s
s= 400/19.6= 20.41m
Thus the maximum height attained is 20.41 m by the ball