Answer:
v = 0
Explanation:
This problem can be solved by taking into account:
- The equation for the calculation of the period in a spring-masss system
( 1 )
- The equation for the velocity of a simple harmonic motion
( 2 )
where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block
Hence

and by reeplacing it in ( 2 ):

In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.
Answer:
8m/s
Explanation:
Average Speed = distance / time = 400/50 = 8m/s
Answer:
<em>The penny will hit the ground at 6.39 seconds</em>
Explanation:
<u>Free Fall</u>
The penny is dropped from a height of y=200 m. The equation of the height on a free-fall motion is given by:

Where
, and t is the time.
Solving for t:

Using the value y=200:

t=6.39 sec
The penny will hit the ground at 6.39 seconds