In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is

where D=5.00 m is the distance of the screen from the slits, and

is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:

And from the relationship between frequency and wavelength,

, we can find the frequency of the light:
The cyclist who travels 20 kilometers per hour for 15 kilometers
There not enough inform here to answer this question
Answer:
a= - 6.667 m/s²
Explanation:
Given that
The initial speed of the box ,u= 20 m/s
The final speed of the box ,v= 0 m/s
The distance cover by box ,s= 30 m
Lets take the acceleration of the box = a
We know that
v²= u ² + 2 a s
Now by putting the values in the above equation we get
0²=20² + 2 a x 30

a= - 6.667 m/s²
Negative sign indicates that velocity and acceleration are in opposite direction.
Therefore the acceleration of the box will be - 6.667 m/s² .