Electrical energy is the starting and final energy of a battery
<span>Weight of the skydiver m = 500 N
Terminal velocity V = 90 km/h
Here the weight of the person acts as the force, so based on the Newton's third law the applied is the force what we but in the opposite direction making the resistance. So the air resistance exerted on Suzie will be her weight that is 500N</span>
Answer:

Explanation:
The formula for potential energy is:

where <em>m </em>is the mass, <em>g</em> is the gravitational acceleration, and <em>h</em> is the height.
The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.

Substitute the values into the formula.

Multiply the first two numbers.
- 0.4 kg*9.8 m/s²= 3.92 kg*m/s²
- If we convert the units now, the problem will be much easier later on.
- 1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N

Multiply.
- 3.92 N* 2 m=7.84 N*m
- 1 Newton meter is equal to 1 Joule (this is why we converted the units).
- Our answer is equal to<u> 7.84 Joules.</u>

Answer:
4km
Explanation:
15 minutes is 1/4 of an hour.
1/4 of 16 is 4.
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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