Answer:
The vibrational frequency of the rope is 5 Hz.
Explanation:
Given;
number of complete oscillation of the rope, n = 20
time taken to make the oscillations, t = 4.00 s
The vibrational frequency of the rope is calculated as follows;
Therefore, the vibrational frequency of the rope is 5 Hz.
The magnitude of the force exerted by the left cube on the right cube is 17.64N.
<h3>What is frictional force?</h3>
When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction.
Two identical 3.0-kg cubes are placed on a horizontal surface in contact with one another. The cubes are lined up from left to right and a force F₁ is applied to the left side of the left cube causing both cubes to move at a constant speed v. The coefficient of kinetic friction between the cubes and the surface is 0.3.
From the equilibrium of forces in vertical direction
Normal force N= 2m x g
friction force f = μN =μ(2m)g
From the equilibrium of forces in horizontal direction
F₁ =ma =0
using Newton's third law of motion, we get
F₁ - f =0
F₁ =f = μ(2m)g
Put the values, we get
F₁ = 17.64N
Thus, magnitude of the force exerted by the left cube on the right cube is 17.64N.
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Explanation:
According to Rydberg's formula, the wavelength of the balmer series is given by:
R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:
Here, 
is the "general" Rydberg constant, 
is electron's mass and M is the mass of the atom nucleus
For hydrogen, we have, 
:
Now, we calculate the wavelength for hydrogen:
![\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%3DR_H%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5C%5C%5Clambda%3D%5BR_H%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D%5B1.0967%2A10%5E7m%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D6.5646%2A10%5E%7B-7%7Dm%3D656.46nm)
For deuterium, we have 
:
![R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm](https://tex.z-dn.net/?f=R_D%3D%5Cfrac%7B1.09737%2A10%5E7m%5E%7B-1%7D%7D%7B%281%2B%5Cfrac%7B9.11%2A10%5E%7B-31%7Dkg%7D%7B2%2A1.67%2A10%5E%7B-27%7Dkg%7D%29%7D%5C%5CR_D%3D1.09707%2A10%5E7m%5E%7B-1%7D%5C%5C%5C%5C%5Clambda%3D%5BR_D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D%5B1.09707%2A10%5E7m%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D6.5629%2A10%5E%7B-7%7D%3D656.29nm)
The scientists measured the amount of carbon dioxide in the atmosphere for dates prior to 1955, From air bubbles in glacial ice cores.
Almost 412 parts per million (ppm) of carbon dioxide are currently present in the Earth's atmosphere, and that number is increasing. This indicates an 11 per cent increase since 2000, when the concentration was close to 370 ppm, and a 47 per cent increase since the start of the Industrial Age when the concentration was close to 280 ppm. Most of the time, scientists do not consider Carbon dioxide as a percentage of the earth's atmosphere. It helps to picture a million equal pieces of an atmospheric gas sample.
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