<span>The density of the energy that is transported under the waves under the ocean surface is about five times higher compared to the wind energy 20 meter (about 65 feet) above. In other words, the amount of energy in a single wave is very high. However it turns out that when they tried to make the high power of the waves to power things in a house it only made it past the first few stages there were very few that made it to some stages. This started taking place in around 2001. hope this helped if not let me know and i could explain more:)
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For this case we have the following measures in the international system:
- <em>For the length:
</em>
It is a unit that is used to measure the length of any object: Meter.
The mass in the international system is denoted by: kilogram. It is used to have an approximate idea of the weight of objects.
The measure in the international system is: Seconds. It is used to know how much time elapses in an event to be studied.
- <em>For the temperature:
</em>
The measure in the international system is: Kelvin. It is used to get an idea of how cold or hot an object, process, place, person can be.
Answer:
Length: meter
Mass: Kilogram
Time: Second
Temperature: Kelvin
Answer:
Force = 10.244 Newtons
b) No of oscillations = 0.88
Explanation:
Since the block executes SHM we can write it's position as function of time as
![x=Asin(\omega t)](https://tex.z-dn.net/?f=x%3DAsin%28%5Comega%20t%29)
ω is the natural frequency of the system
A is the amplitude of the system
![\omega =\sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=%5Comega%20%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
Thus accleration of the block
![x=Asin(\omega t)\\\\a=\frac{d^{2}x(t)}{dt^{2}}\\\\a=\frac{d^{2}Asin(\omega t)}{dt^{2}}\\\\a=-A\omega ^{2}sin(\omega t)](https://tex.z-dn.net/?f=x%3DAsin%28%5Comega%20t%29%5C%5C%5C%5Ca%3D%5Cfrac%7Bd%5E%7B2%7Dx%28t%29%7D%7Bdt%5E%7B2%7D%7D%5C%5C%5C%5Ca%3D%5Cfrac%7Bd%5E%7B2%7DAsin%28%5Comega%20t%29%7D%7Bdt%5E%7B2%7D%7D%5C%5C%5C%5Ca%3D-A%5Comega%20%5E%7B2%7Dsin%28%5Comega%20t%29)
Thus using the given values at t= 3.50 sec we can calculate the acceleration as
![k=\sqrt{\frac{5}{2}}=1.58rad/sec\\\\A=3.0m\\\\a=-3.0\times (1.58)^{2}sin(1.58\times 3.5)\\\\a=5.122m/s^{2}](https://tex.z-dn.net/?f=k%3D%5Csqrt%7B%5Cfrac%7B5%7D%7B2%7D%7D%3D1.58rad%2Fsec%5C%5C%5C%5CA%3D3.0m%5C%5C%5C%5Ca%3D-3.0%5Ctimes%20%281.58%29%5E%7B2%7Dsin%281.58%5Ctimes%203.5%29%5C%5C%5C%5Ca%3D5.122m%2Fs%5E%7B2%7D)
thus force can be calculated using newtons second law as
![Force = m\times accleration\\\\force=2\times 5.122=10.244 Newtons](https://tex.z-dn.net/?f=Force%20%3D%20m%5Ctimes%20accleration%5C%5C%5C%5Cforce%3D2%5Ctimes%205.122%3D10.244%20Newtons)
b)
Now no of oscillations can be obtained as
![\frac{time}{TimePeriod}\\\\Time Period=\frac{2\pi }{w} \\\\time Period=\frac{2\pi }{1.58}\\\\ TimePeriod=3.976secs\\\\](https://tex.z-dn.net/?f=%5Cfrac%7Btime%7D%7BTimePeriod%7D%5C%5C%5C%5CTime%20Period%3D%5Cfrac%7B2%5Cpi%20%7D%7Bw%7D%20%5C%5C%5C%5Ctime%20Period%3D%5Cfrac%7B2%5Cpi%20%7D%7B1.58%7D%5C%5C%5C%5C%20TimePeriod%3D3.976secs%5C%5C%5C%5C)
no of oscillations in 3.50 seconds = 3.50/3.976 = 0.88
What unit? I don’t see anything to help you answer.