Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π ×
................2
put here value
period of oscillations = 2π ×
period of oscillations = 0.6953
so period of oscillations is 0.695 second
Answer:
40m
Explanation:
let's calculate the acceleration first
force = mass × acceleration
rearranging to find acceleration:
acceleration = force ÷ mass
force = 25N, mass = 5.0kg
acceleration = 25 ÷ 5 = 5ms^-2
we can now use the formula v^2 = u^2 + 2as where v = final velocity, u = initial velocity, a = acceleration and s = distance
rearranging v^2 = u^2 + 2as the distance is
s = (v^2 - u^2) ÷ 2a
v = 20, u = 0, a = 5
s = (20^2 - 0^2) ÷ (2 × 5) = 40m
the distance is 40m
Answer:
C. rollback
Explanation:
The retina of the eye is located inside the eye making it impossible for the retina to roll back. Only the eye itself can rollback.
She is traveling at a constant speed.
Answer:
28 miles
Explanation:
3,223 meters/day x 14 days = 45,122 meters
1 meter = 3.28 feet
45,122 meters = 148,000 feet
5280 feet = 1 mile
148,000 feet = 28 miles