The frequency is exactly the rate at which the bug beats its wings.
If that "600" that you mentioned is 600 beats per second, then
THAT's the frequency . . . 600 per second. (600 Hz)
Answer:
The near point of an eye with power of +2 dopters, u' = - 50 cm
Given:
Power of a contact lens, P = +2.0 diopters
Solution:
To calculate the near point, we need to find the focal length of the lens which is given by:
Power, P = 
where
f = focal length
Thus
f = 
f =
= + 0.5 m
The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.
Now, by using lens maker formula:

where
u = object distance = 25 cm = 0.25 m = near point of a normal eye
u' = image distance
Now,



Solving the above eqn, we get:
u' = - 0.5 m = - 50 cm
a)
, 
The work done by the student in each trial is equal to the gravitational potential energy gained by the student:

where
m = 68 kg is the mass of the student
g = 9.8 m/s^2 is the acceleration of gravity
is the gain in height of the student
For the first student,
, so the work done is

The second student runs up to the same height (3.5 m), so the work done by the second student is the same:

2)
, 
The power exerted by each student is given by

where
W is the work done
t is the time taken
For the first student,
and
, so the power exerted is

For the second student,
and
, so the power exerted is
