Answer:
The width of the central bright fringe on the screen is observed to be unchanged is 
Explanation:
To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

Where,
w = width
wavelength
m is an integer, m = 1, 2, 3...
We here know that as
as w are constant, then

We need to find
, then

Replacing with our values:


Therefore the width of the central bright fringe on the screen is observed to be unchanged is 
Middle school? lol but, the third one.
To solve the problem it is necessary to apply conservation of the moment and conservation of energy.
By conservation of the moment we know that

Where
M=Heavier mass
V = Velocity of heavier mass
m = lighter mass
v = velocity of lighter mass
That equation in function of the velocity of heavier mass is

Also we have that 
On the other hand we have from law of conservation of energy that

Where,
W_f = Work made by friction
KE = Kinetic Force
Applying this equation in heavier object.






Here we can apply the law of conservation of energy for light mass, then

Replacing the value of 

Deleting constants,


Answer:
Explanation:
The law of effect given by Edward Thorndike explains the behavior of child given in the problem . The effect which gives pleasure and satisfying effect are likely to be learnt easily and the effect which creates bad taste are likely to be forgotten easily . actions producing feel good effect are likely to be repeated in future .
When the life preserver is dropped from the helicopter, the only force acting on the object is the gravitational force. This modifies the equations of motion. Thus, the working equation is written below:
h = vt + 0.5gt²
where
v is the initial velocity
g is the acceleration due to gravity equal to 9.81 m/s²
h is the height of the fall
h = (1.46 m/s)(1.8 s) + 0.5(9.81 m/s²)(1.8 s)
h = 11.457 m