1. Velocity at which the packet reaches the ground: 121.2 m/s
The motion of the packet is a uniformly accelerated motion, with constant acceleration 
directed downward, initial vertical position 
, and initial vertical velocity 
. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:
substituting, we find
2. height at which the packet has half this velocity: 562.6 m
We need to find the heigth at which the packet has a velocity of
In order to do that, we use again the same SUVAT equation substituting 
with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:
Which means that the heigth of the packet was
Answer:
True.
Explanation:
A nanometer is a unit of mass, whereas a nanosecond is a unit of time. To convert 1.3 hours to minute, you would multiply by 1 h / 60 min. Kilometer is a unit of length, whereas kilogram is a unit of mass. True.
Answer:81.235N
Explanation:
Work=88J
theta=10°
distance=1.1 meters
work=force x cos(theta) x distance
88=force x cos10 x 1.1 cos10=0.9848
88=force x 0.9848 x 1.1
88=force x 1.08328
Divide both sides by 1.08328
88/1.08328=(force x 1.08328)/1.08328
81.235=force
Force=81.235
We have that the spring constant is mathematically given as
Generally, the equation for angular velocity is mathematically given by
Where
k=spring constant
And
Therefore
Hence giving spring constant k
Generally
Mass of earth 
Period for on complete resolution of Earth around the Sun
Therefore
In conclusion
The effective spring constant of this simple harmonic motion is
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t = 0.527 s
<u>It accelerates for 0.527 s.</u>
<u>Explanation:</u>
We use the formula:
v = u+at
Given:
v = 106 m/s
u = 0 (since no gravity)
So applying the formula,
v = u+at
106 = 0 + 201t
t = 106/201
t = 0.527 s
