Answer:
a
b
Explanation:
From the question we are told that
The distance of separation is 
The is distance of the screen from the slit is 
The distance between the central bright fringe and either of the adjacent bright 
Generally the condition for constructive interference is
From the question we are told that small-angle approximation is valid here.
So 
=> 
=> 
Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes
Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as
From the question we are told that small-angle approximation is valid here.
So
=> 
So
substituting values
In the b part of the question we are considering the next set of bright fringe so n= 2
Hence
![\theta = sin^{-1}[\frac{ n * \lambda }{d} ]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7B%20n%20%20%2A%20%20%5Clambda%20%7D%7Bd%7D%20%5D)
![\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20sin%5E%7B-1%7D%5B%5Cfrac%7B%202%20%20%2A%20%201667%20%2A10%5E%7B-9%7D%7D%7B%200.00022%7D%20%5D)
Answer:
The correct answer to this question is D. The area of a desk top.
Explanation:
Square meters, or m^2, are the units that represent area, or the product of two lengths. This is because the length and the width both have the units meters, so when multiplied to calculate area, the resulting units are square meters or m2. Thus, the answer for this question is D.
The length of a pendulum string would be measured in meters.
The volume of an atom would be measured in m^3 (volume is the product of the length, width, and height, giving it a cubic power). However, the volume would probably be measured in a much smaller unit because atoms are very small.
The mass of a book would likely be measured in grams or kilograms.
Answer:
Don't you worry, 'cause everything's gonna be alright, ai-a'ight
Be alright, ai-a'ight
Explanation:
Answer:
Maximum force, F = 1809.55 N
Explanation:
Given that,
Diameter of the anterior cruciate ligament, d = 4.8 mm
Radius, r = 2.4 mm
The tensile strength of the anterior cruciate ligament, 
We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,
So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N
You just need to replace x with 5 in each function
.5^5 - 11
-5-3
.5 ^-6
-8
64 - 8 = 56 A Celcius
Hope this helps
