Hello!
Answer:
Regular physical activity in life improves strength and endurance, helps build healthy bones and muscles, helps control weight, reduces anxiety and stress, increases self-esteem, and it can improve blood pressure and cholesterol levels.
<u></u>
<u>Hope this help<3</u>
Answer:
The speed of the wave is ![f_{1}2L](https://tex.z-dn.net/?f=f_%7B1%7D2L)
Explanation:
Given that,
Frequency = f
String length = l
We need to drive a formula of the speed of the wave
Using formula of fundamental frequency
![f=\dfrac{v}{\lambda}](https://tex.z-dn.net/?f=f%3D%5Cdfrac%7Bv%7D%7B%5Clambda%7D)
We know that,
![\lambda=2L](https://tex.z-dn.net/?f=%5Clambda%3D2L)
![f=\dfrac{v}{2L}](https://tex.z-dn.net/?f=f%3D%5Cdfrac%7Bv%7D%7B2L%7D)
Where, v = speed of wave
= wavelength of the wave
f = frequency
The speed of the wave is
![v=f_{1}2L](https://tex.z-dn.net/?f=v%3Df_%7B1%7D2L)
Hence, The speed of the wave is ![f_{1}2L](https://tex.z-dn.net/?f=f_%7B1%7D2L)
I'm not completely sure, but I think it's 3.4 newtons. I hope you get it correct.
Answer:
Explanation:
what do you really need help with
Answer:
The Gauge pressure at 9 meters depth is ![150 \, kPa](https://tex.z-dn.net/?f=150%20%5C%2C%20kPa)
Explanation:
Gauge pressure is the difference between absolute pressure and some reference pressure, most commonly atmospheric pressure. The increment in pressure caused by a static fluid is given by:
where
is the density of the liquid, g is the accleration due to gravity and d is the depth.
Now, we see that
is linearly proportional to d, and we can assume that
remains constant, because liquids are usually not compressible.
Given that the greater depth is simply 3 times the smaller depth:
![d_2=3\cdot d_1\\9\,m= 3 \cdot 3\,m](https://tex.z-dn.net/?f=d_2%3D3%5Ccdot%20d_1%5C%5C9%5C%2Cm%3D%203%20%5Ccdot%203%5C%2Cm)
at
of depth will also be three times the gauge pressure at
of depth.
We could also have calculated
ny using:
![\Delta P = \rho \,g \,d\\\\\rho = \frac{\Delta P}{g \, d}\\\\\rho = \frac{\Delta P}{g \, d}= \frac{30 \,kPa}{9.8 \frac{m}{s^2} \, 3\,m}=1020.41 \frac{kg}{m^3}](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%5Crho%20%5C%2Cg%20%5C%2Cd%5C%5C%5C%5C%5Crho%20%3D%20%5Cfrac%7B%5CDelta%20P%7D%7Bg%20%5C%2C%20d%7D%5C%5C%5C%5C%5Crho%20%3D%20%5Cfrac%7B%5CDelta%20P%7D%7Bg%20%5C%2C%20d%7D%3D%20%5Cfrac%7B30%20%5C%2CkPa%7D%7B9.8%20%5Cfrac%7Bm%7D%7Bs%5E2%7D%20%20%5C%2C%203%5C%2Cm%7D%3D1020.41%20%5Cfrac%7Bkg%7D%7Bm%5E3%7D)
and used this result to calculate the gauge pressure. These are both similar methods that yield the same result