Answer:
p = 6.64 cm
Explanation:
For this exercise we use the equation of the constructor

where f is the focal length, p and q are the distance to the object and the image, respectively
They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image
1 / p = 1 / 2.2 - 1/3.29
1 / p = 0.15059
p = 6.64 cm
therefore the farthest distance from the object is 6.64 c
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The statement: Mass affects how fast an object falls is true.
Answer:
x = A sin ω t describes the displacement of the particle
v = A ω cos ω t
a = -A ω^2 sin ω t
a (max) = -A ω^2 is the max acceleration (- can be ignored here)
ω = (K/ m)^1/2 for SHM
F = - K x^2 restoring force of spring
K = 4.34 / .0745^2 = 782 N / m
ω = (782 / .297)^1/2 = 51.3 / sec
a (max) = .0745 * 782 / .297 = 196 m / s^2