The periods of oscillation for the mass–spring systems from largest to smallest is:
- m = 4 kg , k = 2 N/m (T = 8.89 s)
- m = 2 kg , k = 2 N/m (T = 6.28 s)
- m = 2 kg , k = 4 N/m (T = 4.44 s)
- m = 1 kg , k = 4 N/m (T = 3.14 s)
<h3>Explanation:</h3>
The period of oscillation in a simple harmonic motion is defined as the following formulation:

Where:
T = period of oscillation
m = inertia mass of the oscillating body
k = spring constant
m = 2 kg , k = 2 N/m


T = 6.28 s
m = 2 kg , k = 4 N/m


T = 4.44 s
m = 4 kg , k = 2 N/m


T = 8.89 s
m = 1 kg , k = 4 N/m


T = 3.14 s
Therefore the rank the periods of oscillation for the mass–spring systems from largest to smallest is:
- m = 4 kg , k = 2 N/m (T = 8.89 s)
- m = 2 kg , k = 2 N/m (T = 6.28 s)
- m = 2 kg , k = 4 N/m (T = 4.44 s)
- m = 1 kg , k = 4 N/m (T = 3.14 s)
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Answer:
Mercury / Mars
Explanation:
For an object launched straight upward, the following SUVAT equation can be used

where
v is the final velocity
u is the initial velocity
g is the acceleration of gravity (free fall acceleration) (the negative sign is due to the downward direction of gravity)
h is the maximum height reached
At the maximum height, the velocity is zero, so v = 0. Re-arranging the equation,

So we see that for equal initial velocity (u), the maximum height reaches is inversely proportional to the acceleration of gravity. Therefore, the potato gun will reach the highest altitude in the planets with lowest acceleration of gravity, therefore Mercury and Mars (3.7 and 3.6 m/s^2).
Answer:
(a): The car's relative position to the base of the cliff is x= 32.52m.
(b): The lenght of the car in the ir is tfall= 1.78 sec.
Explanation:
Vo= 0
V= ?
d= 50m
h= 30m
a= 4 m/s²
t= √(2*d/a)
t= 5 sec
V= a*t
V= 20 m/s
Vx= V * cos(24º)
Vx= 18.27 m/s
Vy= V* sin(24º)
Vy= 8.13 m/s
h= Vy*t + g*t²/2
clearing t:
tfall= 1.78 sec (b)
x= Vx * tfall
x= 32.52 m (a)
Transverse waves are always characterized by particle motion being perpendicular to wave motion. A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves.
Answer:

Acceleration, in m/s, of such a rock fragment = 
Explanation:
According to Newton's Third Equation of motion

Where:
is the final velocity
is the initial velocity
a is the acceleration
s is the distance
In our case:

So Equation will become:

Acceleration, in m/s, of such a rock fragment = 