The answer is false your welcome
Answer:
![r_{cm}=[12.73,12.73]cm](https://tex.z-dn.net/?f=r_%7Bcm%7D%3D%5B12.73%2C12.73%5Dcm)
Explanation:
The general equation to calculate the center of mass is:
Any differential of mass can be calculated as:
Where "a" is the radius of the circle and λ is the linear density of the wire.
The linear density is given by:
So, the differential of mass is:
Now we proceed to calculate X and Y coordinates of the center of mass separately:
Solving both integrals, we get:
Therefore, the position of the center of mass is:
The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
Read more about vertical weight here:
brainly.com/question/15244771
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Your question has been heard loud and clear.
Well it depends on the magnitude of charges. Generally , when both positive charges have the same magnitude , their equilibrium point is towards the centre joining the two charges. But if magnitude of one positive charge is higher than the other , then the equilibrium point will be towards the charge having lesser magnitude.
Now , a negative charge is placed in between the two positive charges. So , if both positive charges have same magnitude , they both pull the negative charge towards each other with an equal force. Thus the equilibrium point will be where the negative charge is placed because , both forces are equal , and opposite , so they cancel out each other at the point where the negative charge is placed. However if they are of different magnitudes , then the equilibrium point will be shifted towards the positive charge having less magnitude.
Thank you
Answer:
Explanation:
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. <em>FALSE. </em>The specific lines are obseved because of the energy level transition of an electron in an specific level to another level of energy.
The energies of atoms are not quantized. <em>FALSE. </em>The energies of the atoms are in specific levels.
When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. <em>FALSE. </em>During absorption, a specific wavelength of light is absorbed, not emmited.
Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. <em>TRUE. </em>Again, you can observe just the transition due the change of energy of an electron in the quantized energy level
When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. <em>TRUE. </em>The electron decreases its energy releasing a specific wavelength of light.
The energies of atoms are quantized. <em>TRUE. </em>In fact, the energy of all subatomic, atomic, and molecular particles is quantized.
