Answer:
Explanation:
Given
Initial Intensity of light is S
when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.
When it is passed through a second Polarizer with its transmission axis 

here 


When it is passed through third Polarizer with its axis
to first but
to second thus 



When middle sheet is absent then Final Intensity will be zero
Answer: C
Explanation: The correct answer would be Inertial. I take the quiz and got the answer rght!!
A closed circle means the number is included and an open circle means its not.
<u>Answer:</u>
2N/cm
<u>Step-by-step explanation:</u>
According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

where,
is the force which is stretching or compressing the spring,
is the spring constant; and
is the distance the spring is stretched.
Substituting the given values to find the elastic constant
to get:




Therefore, the elastic constant is 2 Newton/cm.