D.all of the above is the answer for this question
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
i will give you 25 points sorry becuase i can't help with that
The absolute magnitude of the star would be +5.
Answer:
310 meters
Explanation:
Given:
v₀ = 0 m/s
t = 8.0 s
a = -9.8 m/s²
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (8.0 s) + ½ (-9.8 m/s²) (8.0 s)²
Δy = -313.6
Rounded to two significant figures, the object fell 310 meters.