Answer:
Her centripetal acceleration during the turn at each end of the track is 
Step-by-step explanation:
Total distance covered in one round , D= 400 m
Time taken to cover one round , T = 1 min 40 s = 100 sec
Speed of runner , 
Now centripetal acceleration is given by
where 
Thus her centripetal acceleration during the turn at each end of the track is
Your answer from smallest to largest is:
3/10
3/5
13/20
22/25
(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))
Recall that
tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)
so cos²(<em>θ</em>) cancels with the cos²(<em>θ</em>) in the tan²(<em>θ</em>) term:
(sin²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))
Recall the double angle identity for cosine,
cos(2<em>θ</em>) = 2 cos²(<em>θ</em>) - 1
so the 1 in the denominator also vanishes:
(sin²(<em>θ</em>) - 1) / (2 cos²(<em>θ</em>))
Recall the Pythagorean identity,
cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1
which means
sin²(<em>θ</em>) - 1 = -cos²(<em>θ</em>):
-cos²(<em>θ</em>) / (2 cos²(<em>θ</em>))
Cancel the cos²(<em>θ</em>) terms to end up with
(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>)) = -1/2
fg^3 / g^3f = 1
f^4 / f*f*f*f = f^4 / f^4 = 1
f^2g^3 / f^3g^2 = g/f
f*f/f^2 = f^2 / f^2 = 1
Answer:
f^2g^3 / f^3g^2
