Answer: A plz brainliest me
An isotope is one of two or more forms of the same chemical element. Different isotopes of an element have the same number of protons in the nucleus, giving them the same atomic number, but a different number of neutrons giving each elemental isotope a different atomic weight.
Explanation:
Answer:
(C) 16 radians
Explanation:
The angular displacement is given by the following equation:
![\Delta \theta=\omega_i t+\frac{1}{2}\alpha t^2](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%3D%5Comega_i%20t%2B%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2)
Here
Is the angular displacement of the body at the indicated time (t).
Is the angular velocity of the body at the initial moment.
Is the angular acceleration of the body.
The disk starts from rest, so ![\omega_i=0](https://tex.z-dn.net/?f=%5Comega_i%3D0)
Replacing the given values:
![\Delta \theta=\frac{1}{2}(2\frac{radians}{s^2})(4s)^2\\\Delta \theta=16 radians](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%3D%5Cfrac%7B1%7D%7B2%7D%282%5Cfrac%7Bradians%7D%7Bs%5E2%7D%29%284s%29%5E2%5C%5C%5CDelta%20%5Ctheta%3D16%20radians)
If you are referring to stars, the answer would then be pressure from the nuclear reactions
The thermal pressure that pushes outward and against the pull of gravity in a star is caused by the nuclear reactions that is happening within the stars core. A lot of energy is released during these reactions which produce thermal pressure. The pressure then pushes outward.
A chemical change might also result in the formation of a precipitate, such as the appearance of a cloudy material when dissolved substances are mixed. Rotting, burning, cooking, and rusting are all further types of chemical changes because they produce substances that are entirely new chemical compounds.
You are working for a manufacturing company, which is mathematically given as
<h3>What is the
value of m that will place the
movable bead in equilibrium at x-a a ....?</h3>
a)
Generally, the equation for the force of equilibrium is mathematically given as
F=2fcos\theta
Therefore
![K(npq^2/a^2)=2\frac{kmpq^2}{a\sqrt{2}^2}0.5\\\\np=mP/ \sqrt{2}\\\\where n=3](https://tex.z-dn.net/?f=K%28npq%5E2%2Fa%5E2%29%3D2%5Cfrac%7Bkmpq%5E2%7D%7Ba%5Csqrt%7B2%7D%5E2%7D0.5%5C%5C%5C%5Cnp%3DmP%2F%20%5Csqrt%7B2%7D%5C%5C%5C%5Cwhere%20n%3D3)
![m=3\sqrt{2}](https://tex.z-dn.net/?f=m%3D3%5Csqrt%7B2%7D)
b)
By force equilibrium
![K(npq^2/(2a^2))=2*\frac{kmpq^2}{a\sqrt{5a}^2}* \frac{2a}{\sart{5a}}](https://tex.z-dn.net/?f=K%28npq%5E2%2F%282a%5E2%29%29%3D2%2A%5Cfrac%7Bkmpq%5E2%7D%7Ba%5Csqrt%7B5a%7D%5E2%7D%2A%20%5Cfrac%7B2a%7D%7B%5Csart%7B5a%7D%7D)
Therefore
![n/4=2/5*m*2/\sqrt{5}\\\\m= \frac{5\sqrt{5}}{16}\\\\\\\\](https://tex.z-dn.net/?f=n%2F4%3D2%2F5%2Am%2A2%2F%5Csqrt%7B5%7D%5C%5C%5C%5Cm%3D%20%5Cfrac%7B5%5Csqrt%7B5%7D%7D%7B16%7D%5C%5C%5C%5C%5C%5C%5C%5C)
![m=\frac{15\sqrt{5}}{16}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B15%5Csqrt%7B5%7D%7D%7B16%7D)
c)
![K(npq^2/x^2)=2\frac{kmpq^2}{a\sqrt{x^2+a^2}^2}0.5*x/\sqrt{x^2+a^2}](https://tex.z-dn.net/?f=K%28npq%5E2%2Fx%5E2%29%3D2%5Cfrac%7Bkmpq%5E2%7D%7Ba%5Csqrt%7Bx%5E2%2Ba%5E2%7D%5E2%7D0.5%2Ax%2F%5Csqrt%7Bx%5E2%2Ba%5E2%7D)
x^2+a^2=(14/3)^{2/3}x^2
x=a/1.338
x=0.747a
d)
By force equilibrium
![K(npq^2/x^2)=2\frac{kmpq^2n}{\sqrt{x^2+a^{3/2}}^2}\\\\n/x^2=\frac{2mx}{(x^2+a^2)^{3/2}}](https://tex.z-dn.net/?f=K%28npq%5E2%2Fx%5E2%29%3D2%5Cfrac%7Bkmpq%5E2n%7D%7B%5Csqrt%7Bx%5E2%2Ba%5E%7B3%2F2%7D%7D%5E2%7D%5C%5C%5C%5Cn%2Fx%5E2%3D%5Cfrac%7B2mx%7D%7B%28x%5E2%2Ba%5E2%29%5E%7B3%2F2%7D%7D)
![m/n=\frac{(x^2+a^2)3/2}{x^3}](https://tex.z-dn.net/?f=m%2Fn%3D%5Cfrac%7B%28x%5E2%2Ba%5E2%293%2F2%7D%7Bx%5E3%7D)
Read more about electric fields
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