Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm
Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as
y = R×(mλ/d)
Where y = distance between nth fringes and the center fringe.
m = order of fringe
λ = wavelength of light = 589nm = 589×10^-9m
R = distance between slits and screen = 1.0m
d = distance between slits = 0.25mm = 0.00025m
For distance between the first dark fringe and the center fringe.
This implies that m = 1
y = 1 × 589×10^-9 × 1/0.00025
y = 589×10^-9/0.00025
y = 2,356,000 × 10^-9
y = 2.356 × 10^-3 m = 2.356mm
For the second dark fringe, this implies that m = 2
y = 1 × 2 × 589×10^-9/0.00025
y = 1178 × 10^-9 /0.00025
y = 4,712,000 × 10^-9
y = 4.712 × 10^-3 m = 4.712mm
Answer:
Part a)
Part b)
Direction = upwards
Explanation:
When ball is dropped from height h = 4.0 m
then the speed of the ball just before it will strike the ground is given as
Now ball will rebound to height h = 2.00 m
so the velocity of ball just after it will rebound is given as
Part a)
Average acceleration is given as
Part B)
As we know that ball rebounds upwards after collision while before collision it is moving downwards
So the direction of the acceleration is vertically upwards
Answer:
The question clearly describes the circular motion.
The circular motion equation is
The path of the particle is circular.
Explanation:
In circular motion, the radial acceleration is always towards the center and constant in magnitude. Furthermore, the velocity of the circular motion is always tangential to the circle, that is it is always perpendicular to the radius, hence the acceleration.
Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)
