P always P because P is an awkward Dorian letter that can always be trusted
Given that,
Mass of ring = m
Mass of sphere = M
Radius = R
Distance = √8R
We need to calculate the intensity of gravitational field
Using formula of intensity
![E_{g}=\dfrac{Gmx}{\sqrt{(r^2+x^2)^\frac{3}{2}}}](https://tex.z-dn.net/?f=E_%7Bg%7D%3D%5Cdfrac%7BGmx%7D%7B%5Csqrt%7B%28r%5E2%2Bx%5E2%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%7D)
Put the value into the formula
![E_{g}=\dfrac{Gm\sqrt{8}R}{(R^2+8R^2)^{\frac{3}{2}}}](https://tex.z-dn.net/?f=E_%7Bg%7D%3D%5Cdfrac%7BGm%5Csqrt%7B8%7DR%7D%7B%28R%5E2%2B8R%5E2%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
![E_{g}=\dfrac{2\sqrt{2}Gm}{(9R^2)^{\frac{3}{2}}}](https://tex.z-dn.net/?f=E_%7Bg%7D%3D%5Cdfrac%7B2%5Csqrt%7B2%7DGm%7D%7B%289R%5E2%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
![E_{g}=\dfrac{2\sqrt{2}Gm}{27R^2}](https://tex.z-dn.net/?f=E_%7Bg%7D%3D%5Cdfrac%7B2%5Csqrt%7B2%7DGm%7D%7B27R%5E2%7D)
We need to calculate the force of attraction between the ring and the sphere
Using formula of attraction force
![F=M\times E_{g}](https://tex.z-dn.net/?f=F%3DM%5Ctimes%20E_%7Bg%7D)
Where, M = mass of sphere
E = intensity of gravitational field
Put the value into the formula
![F=\dfrac{2\sqrt{2}GmM}{27R^2}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B2%5Csqrt%7B2%7DGmM%7D%7B27R%5E2%7D)
Hence, The force of attraction between the ring and the sphere is ![\dfrac{2\sqrt{2}GmM}{27R^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%5Csqrt%7B2%7DGmM%7D%7B27R%5E2%7D)
We have that the time, to the nearest minute, when the water level is at 1.125 m for the second time after midnight is
![t=10.0hours](https://tex.z-dn.net/?f=t%3D10.0hours)
From the Question we are told that
Maximum height ![h_{max}=3m](https://tex.z-dn.net/?f=h_%7Bmax%7D%3D3m)
Minimum height ![H_{min}=0.5m](https://tex.z-dn.net/?f=H_%7Bmin%7D%3D0.5m)
Time for next high tide will occur![T=12 hours =>720 min](https://tex.z-dn.net/?f=T%3D12%20hours%20%3D%3E720%20min)
Generally Average Height
![h_{avg}=\frac{3+0.5}{2}\\\\h_{avg}=1.75](https://tex.z-dn.net/?f=h_%7Bavg%7D%3D%5Cfrac%7B3%2B0.5%7D%7B2%7D%5C%5C%5C%5Ch_%7Bavg%7D%3D1.75)
Therefore determine Amplitude to be
![A=h_{max}=j_{avg}\\\\A=3-1.75\\\\A=1.25](https://tex.z-dn.net/?f=A%3Dh_%7Bmax%7D%3Dj_%7Bavg%7D%5C%5C%5C%5CA%3D3-1.75%5C%5C%5C%5CA%3D1.25)
Generally, the equation for Time is mathematically given by
At t=0
![h(x)=Acos(Bx)+h_{avg}](https://tex.z-dn.net/?f=h%28x%29%3DAcos%28Bx%29%2Bh_%7Bavg%7D)
Where
![B=\frac{2\pi}{P}\\\\B=\frac{2\pi}{720}\\\\B=8.73*10^{-3}](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B2%5Cpi%7D%7BP%7D%5C%5C%5C%5CB%3D%5Cfrac%7B2%5Cpi%7D%7B720%7D%5C%5C%5C%5CB%3D8.73%2A10%5E%7B-3%7D)
Therefore
![h(t)=Acos8.73*10^{-3}(t)+h_{avg}](https://tex.z-dn.net/?f=h%28t%29%3DAcos8.73%2A10%5E%7B-3%7D%28t%29%2Bh_%7Bavg%7D)
Hence the Time at
is
![1.125(t)=1.25cos(8.73*10^{-3})(t)+1.75](https://tex.z-dn.net/?f=1.125%28t%29%3D1.25cos%288.73%2A10%5E%7B-3%7D%29%28t%29%2B1.75)
![-0.1249t=1.75](https://tex.z-dn.net/?f=-0.1249t%3D1.75)
![t=10.0hours](https://tex.z-dn.net/?f=t%3D10.0hours)
For more information on this visit
brainly.com/question/22361343
<h3><u>Answer;</u> </h3>
= 0.03066 m
<h3><u>Explanation;</u></h3>
Work done or energy by a spring or an elastic material is given by the formula;
Potential energy = 1/2 kx²
Where, k is the spring constant and x is the distance stretched.
Therefore;
47 J = 1/2 (100000 n/m) x²
47 J = 50, 000 x²
x² = 47 /50,000
x² = 0.00094
x = 0.03066 m
Therefore,<u> x = 0.03066 m </u>
<span>a) The law of conservation of momentum is satisfied because there must be air (and other combustion byproducts) expelled behind the aircraft as it moves forward. The collective momentum of those gases is equal in magnitude and opposite in direction to the forward momentum of the aircraft.
b) The aircraft obeys conservation of energy because it is simply converting energy to different forms. In its stationary form, there existed latent chemical energy in both the fuel on board and the oxygen in the air. As the aircraft moves, some of that chemical energy is converted into kinetic energy and heat, but it is indeed conserved.</span>