The number of protons for an atom with 20 neutrons and 20 electrons will be 20.
<h3>
What is an atom ?</h3>
An atom is the smallest particle of an element. It consist of proton, neutron and electron.
To determine the number of protons and also identify the correct symbol for an atom with 20 neutrons and 20 electrons, we must first acknowledge the following facts
- Atom in a neutral state, the number of protons must be equal to number of electrons.
- Atom in a nuclear stable state, number of protons must be equal to the number of neutrons
Therefore, the number of protons for an atom with 20 neutrons and 20 electrons will be 20.
The symbol for this atom is Ca which is Calcium.
Learn more about Atom here: brainly.com/question/6258301
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Answer:
option (c)
Explanation:
Fundamental frequency of segment A = f
Second harmonic frequency of B = fundamental frequency of A .
Tension in both the wires is same and the mass density is also same as the wires are identical.
fundamental frequency of wire A is given by
.... (1)
Second harmonic of B is given by
.... (2)
Equation (1) is equal to equation (2), we get
![\frac{1}{2L_{A}}=\frac{2}{2L_{B}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2L_%7BA%7D%7D%3D%5Cfrac%7B2%7D%7B2L_%7BB%7D%7D)
![L_{B}=2L_{A}](https://tex.z-dn.net/?f=L_%7BB%7D%3D2L_%7BA%7D)
So, LB = 2 L
Thus, the length of wire segment B is 2 times the length of wire segment A.
Birthdays or age? maybe how old it looks
Answer:
Normal force will be equal to 8.945 N
Explanation:
We have given mass of the cylinder m = 8.15 kg
Diameter d = 15 cm
So radius ![r=\frac{d}{2}=\frac{15}{2}=7.5cm = 0.075 m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bd%7D%7B2%7D%3D%5Cfrac%7B15%7D%7B2%7D%3D7.5cm%20%3D%200.075%20m)
Initial angular velocity ![\omega _i=240rpm=\frac{2\times 3.14\times 240}{60}=25.12rad/sec](https://tex.z-dn.net/?f=%5Comega%20_i%3D240rpm%3D%5Cfrac%7B2%5Ctimes%203.14%5Ctimes%20240%7D%7B60%7D%3D25.12rad%2Fsec)
As the cylinder finally comes to rest so final angular velocity ![\omega _f=0rad/sec](https://tex.z-dn.net/?f=%5Comega%20_f%3D0rad%2Fsec)
Before coming to rest cylinder covers a distance of ![\Theta =5.20revolution=5.20\times 2\times 3.14=32.656rad](https://tex.z-dn.net/?f=%5CTheta%20%3D5.20revolution%3D5.20%5Ctimes%202%5Ctimes%203.14%3D32.656rad)
From third equation of motion ![\omega _f^2=\omega_i^2+2\alpha \Theta](https://tex.z-dn.net/?f=%5Comega%20_f%5E2%3D%5Comega_i%5E2%2B2%5Calpha%20%5CTheta)
![0^2=25.12^2-2\times \alpha \times32.656](https://tex.z-dn.net/?f=0%5E2%3D25.12%5E2-2%5Ctimes%20%5Calpha%20%5Ctimes32.656)
![\alpha =9.66rad/sec^2](https://tex.z-dn.net/?f=%5Calpha%20%3D9.66rad%2Fsec%5E2)
Coefficient of kinetic friction ![\mu _k=0.33](https://tex.z-dn.net/?f=%5Cmu%20_k%3D0.33)
Moment of inertia of the solid cylinder ![I=\frac{1}{2}mr^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B1%7D%7B2%7Dmr%5E2)
We know that ![\tau =I\alpha](https://tex.z-dn.net/?f=%5Ctau%20%3DI%5Calpha)
![F\times r =(\frac{1}{2}mr^2)\times \alpha](https://tex.z-dn.net/?f=F%5Ctimes%20r%20%3D%28%5Cfrac%7B1%7D%7B2%7Dmr%5E2%29%5Ctimes%20%5Calpha)
![F =(\frac{1}{2}mr)\times \alpha](https://tex.z-dn.net/?f=F%20%20%3D%28%5Cfrac%7B1%7D%7B2%7Dmr%29%5Ctimes%20%5Calpha)
![F=\frac{1}{2}\times 8.15\times 0.075\times 9.66=2.952N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%208.15%5Ctimes%200.075%5Ctimes%209.66%3D2.952N)
So normal force will be equal to ![N=\frac{F}{\mu _k}=\frac{2.952}{0.33}=8.945N](https://tex.z-dn.net/?f=N%3D%5Cfrac%7BF%7D%7B%5Cmu%20_k%7D%3D%5Cfrac%7B2.952%7D%7B0.33%7D%3D8.945N)
i think its a, good luck on your test