Answer:
0.67 s
Explanation:
This is a simple harmonic motion (SHM).
The displacement,
, of an SHM is given by

A is the amplitude and
is the angular frequency.
We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or
radian.
From trigonometry,
if A and B are complementary.
At
, 


So

At
, 





The period,
, is related to
by

Answer:
It would crack.
Explanation: The pressure from dropping it would crush the eggshell therefore breaking the egg.
The Answer Is A
Hope This Helps !
Explanation:
It is given that,
Velocity in East, 
Velocity in North, 
(a) The resultant velocity is given by :

(b) The width of the river is, d = 80 m
Let t is the time taken by the boat to travel shore to shore. So,


t = 16 seconds
(c) Let x is the distance covered by the boat to reach the opposite shore. So,


x = 48 meters
Hence, this is the required solution.
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