Starting with the equation for newtons second law of motion:
<em>Answer:</em>
<em>When </em><em>a </em><em>body </em><em>is </em><em>moving </em><em>on </em><em>a </em><em>circle </em><em>it </em><em>is </em><em>accelerating </em><em>because </em><em>centripetal </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>it </em><em>towards </em><em>the </em><em>center.</em>
<em>Please </em><em>see</em><em> the</em><em> attached</em><em> picture</em><em>.</em><em>.</em><em>.</em>
<em>From </em><em>the </em><em>above </em><em>diagram,</em><em>we </em><em>can </em><em>say </em><em>the </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>the </em><em>body </em><em>when </em><em>it </em><em>moves </em><em>in </em><em>a </em><em>circle.</em>
<em>Hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
The energy equivalent of 5.0 kg mass would be <span>4.5x10^17.</span>
The answer for the question is D.
Answer:A)0.5 and 2 mm
Explanation:
Given
radius R of wire is 1 mm
magnetic Field at edge(surface) ![=10^{-5} T](https://tex.z-dn.net/?f=%3D10%5E%7B-5%7D%20T)
Magnetic Field at a distance r' from Center is ![B'=5\times 10^{-6} T](https://tex.z-dn.net/?f=B%27%3D5%5Ctimes%2010%5E%7B-6%7D%20T)
and we know
![B=\frac{\mu _0I}{2\pi r}](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%5Cmu%20_0I%7D%7B2%5Cpi%20r%7D)
where ![I= current](https://tex.z-dn.net/?f=I%3D%20current%20)
Permeability of free space
distance from center
For ![r=R](https://tex.z-dn.net/?f=r%3DR)
---1
For ![r=r'](https://tex.z-dn.net/?f=r%3Dr%27)
---2
Divide 1 and 2
![\frac{10^{-5}}{0.5\times 10^{-5}}=\frac{r'}{R}](https://tex.z-dn.net/?f=%5Cfrac%7B10%5E%7B-5%7D%7D%7B0.5%5Ctimes%2010%5E%7B-5%7D%7D%3D%5Cfrac%7Br%27%7D%7BR%7D)
![r'=2 R=2 mm](https://tex.z-dn.net/?f=r%27%3D2%20R%3D2%20mm%20)
If r is inside the wire then
![B=\frac{\mu _0rI}{2\pi R^2}](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%5Cmu%20_0rI%7D%7B2%5Cpi%20R%5E2%7D)
for ![r=R](https://tex.z-dn.net/?f=r%3DR)
---3
for ![r=r"](https://tex.z-dn.net/?f=r%3Dr%22)
----4
Divide 3 and 4
![2=\frac{R}{r"}](https://tex.z-dn.net/?f=2%3D%5Cfrac%7BR%7D%7Br%22%7D)
![r"=0.5 mm](https://tex.z-dn.net/?f=r%22%3D0.5%20mm)