During the parts in the orbit where the moon is farthest away from the earth the tides will be low. Whereas during the parts where the moon is closer to the earth the tides will be higher.
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When the concrete structure avails it maximum strength, it is safer to place construction load on it. Usually, the concrete gain it maximum strength after 28 days. Concrete when cast take initial settlement on 7 days. Curing is done continuously through the period so that there is no crack occur due to heat transfer. There are various method of concrete curing. On a horizontal surface/ roof/ floor, curing is done through ponding or spreading wet sheets. Testiment/ sample of the concrete are cured by immersion in water tank. After 28 days of curing, concrete gains it maximum strength and can be used for test or usage. Construction load limit is considered in designing/ proportioning the specimen. So the applied load is equal or less than the limit for which it is actually designed.
Explanation:
Wavelength in an emission spectrum,
The energy of an electron is given by :
![E=\dfrac{hc}{\lambda}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bhc%7D%7B%5Clambda%7D)
Where
h is the Planck's constant
c is the speed of light
For 435 nm, the energy of the electron will be :
![E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8\ m/s}{435\times 10^{-9}}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B6.63%5Ctimes%2010%5E%7B-34%7D%5Ctimes%203%5Ctimes%2010%5E8%5C%20m%2Fs%7D%7B435%5Ctimes%2010%5E%7B-9%7D%7D)
![E=4.57\times 10^{-19}\ J](https://tex.z-dn.net/?f=E%3D4.57%5Ctimes%2010%5E%7B-19%7D%5C%20J)
We know that ![1\ eV=1.6\times 10^{-19}\ J](https://tex.z-dn.net/?f=1%5C%20eV%3D1.6%5Ctimes%2010%5E%7B-19%7D%5C%20J)
So, ![E=\dfrac{4.57\times 10^{-19}}{1.6\times 10^{-19}}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B4.57%5Ctimes%2010%5E%7B-19%7D%7D%7B1.6%5Ctimes%2010%5E%7B-19%7D%7D)
So, E = 2.86 eV
The energy of the electron dropping from one energy level is 2.86 eV. We know that,
![\dfrac{hc}{\lambda}=E_{n_2}-E_{n_1}](https://tex.z-dn.net/?f=%5Cdfrac%7Bhc%7D%7B%5Clambda%7D%3DE_%7Bn_2%7D-E_%7Bn_1%7D)
From the given energy levels :
![E_5-E_2=-0.544-(-3.403)](https://tex.z-dn.net/?f=E_5-E_2%3D-0.544-%28-3.403%29)
So, the transition must be from E₅ to E₂. Hence, this is the required solution.
Answer:
![c=9\ m/s^2](https://tex.z-dn.net/?f=c%3D9%5C%20m%2Fs%5E2)
Explanation:
It is given that,
Mass of the particle, m = 2 kg
Force acting on the particle, F = -36 N (negative axis)
The position of the particle as a function of time t is given by :
![x=3m+4t+ct^2-2t^3](https://tex.z-dn.net/?f=x%3D3m%2B4t%2Bct%5E2-2t%5E3)
Velocity, ![v=\dfrac{dx}{dt}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bdx%7D%7Bdt%7D)
![v=\dfrac{d(3m+4t+ct^2-2t^3)}{dt}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bd%283m%2B4t%2Bct%5E2-2t%5E3%29%7D%7Bdt%7D)
![v=4+2ct-6t^2](https://tex.z-dn.net/?f=v%3D4%2B2ct-6t%5E2)
Acceleration, ![a=\dfrac{dv}{dt}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bdv%7D%7Bdt%7D)
![a=2c-12t](https://tex.z-dn.net/?f=a%3D2c-12t)
At t = 3 s
![a=(2c-36)\ m/s^2](https://tex.z-dn.net/?f=a%3D%282c-36%29%5C%20m%2Fs%5E2)
Force acting on the particle is given by :
F = ma
![-36\ N=2\ kg\times (2c-36)](https://tex.z-dn.net/?f=-36%5C%20N%3D2%5C%20kg%5Ctimes%20%282c-36%29)
On solving above equation,
. It has a unit same as acceleration.
Hence, this is the required solution.