Explanation:
An object is attached to the spring and then released. It begins to oscillate. If it is displaced a distance 0.125 m from its equilibrium position and released with zero initial speed. The amplitude of a wave is the maximum displacement of the particle. So, its amplitude is 0.125 m.
After 0.800 seconds, its displacement is found to be a distance 0.125 m on the opposite side. The time period will be, 
We know that the relation between the time period and the time period is given by :
f = 0.625 Hz
So, the frequency of the object is 0.625 Hz. Hence, this is the required solution.
I think it's 6/8, you multiply by 2. 3/4 x 2 = 6/8.
B.) Action and reaction forces helps during lift off of a rocket.
Hope this helps!
Answer:
r = 4.21 10⁷ m
Explanation:
Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining
T² = (
) r³ (1)
in this case the period of the season is
T₁ = 93 min (60 s / 1 min) = 5580 s
r₁ = 410 + 6370 = 6780 km
r₁ = 6.780 10⁶ m
for the satellite
T₂ = 24 h (3600 s / 1h) = 86 400 s
if we substitute in equation 1
T² = K r³
K = T₁²/r₁³
K = 
K = 9.99 10⁻¹⁴ s² / m³
we can replace the satellite values
r³ = T² / K
r³ = 86400² / 9.99 10⁻¹⁴
r = ∛(7.4724 10²²)
r = 4.21 10⁷ m
this distance is from the center of the earth
The answer is <span>3688.25 cm^2</span>
