Answer:
a)F=698.83 N
b)K=8221.56 N/m
Explanation:
Given that
mass ,m = 0.12 kg
Amplitude ,A= 8.5 cm
time period ,T = 0.2 s
We know that
We know that
K=Spring constant
K=8221.56 N/m
The maximum force F
F= K A
F= 8221.56 x 0.085 N
F=698.83 N
a)F=698.83 N
b)K=8221.56 N/m
The distance between slit and the screen is 1.214m.
To find the answer, we have to know about the width of the central maximum.
<h3>How to find the distance between slit and the screen?</h3>
where, d is the distance between slit and the screen, and a is the slit width.
Thus, we can conclude that, the distance between slit and the screen is 1.214m.
Learn more about the width of the central maximum here:
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