Answer:
Newton, symbol N
Explanation:
It is the newton, whose symbol is N.
Answer:
is proved.
Explanation:
The magnetic field in the long current carrying wire is,
![B=\frac{\mu_{0}I }{2\pi r } \phi](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%5Cmu_%7B0%7DI%20%7D%7B2%5Cpi%20r%20%7D%20%5Cphi)
Here, I is the current, B is the magnetic field.
Now, by using cylindrical coordinates for the divergence of B.
![\bigtriangledown.B=\frac{1}{s} \frac{d}{d\phi} B](https://tex.z-dn.net/?f=%5Cbigtriangledown.B%3D%5Cfrac%7B1%7D%7Bs%7D%20%5Cfrac%7Bd%7D%7Bd%5Cphi%7D%20B)
Put the value of B in above equation.
![\bigtriangledown.B=\frac{1}{s} \frac{d}{d\phi}(\frac{\mu_{0}I }{2\pi r } \phi)\\\bigtriangledown.B=0](https://tex.z-dn.net/?f=%5Cbigtriangledown.B%3D%5Cfrac%7B1%7D%7Bs%7D%20%5Cfrac%7Bd%7D%7Bd%5Cphi%7D%28%5Cfrac%7B%5Cmu_%7B0%7DI%20%7D%7B2%5Cpi%20r%20%7D%20%5Cphi%29%5C%5C%5Cbigtriangledown.B%3D0)
Hence, it is prove that for a long current I carrying wire magnetic field divergence that is
.
<span>The frequency is defined as the number of cycles done in one second:
</span>
![f= \frac{N}{t}](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7BN%7D%7Bt%7D%20)
<span>
where N is the number of cycles and t is the time taken to complete N cycles. The wheel in this problem does N=2 revolutions in a time of t=4.0s, therefore its frequency of rotation is:
</span>
![f= \frac{N}{t} = \frac{2.0}{4.0s} = 0.5 Hz](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7BN%7D%7Bt%7D%20%3D%20%5Cfrac%7B2.0%7D%7B4.0s%7D%20%20%3D%200.5%20Hz)
<span>
</span>
Answer:
Explanation:
1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
For constant acceleration along a given direction, we can relate acceleration, velocity and position with the following equation that doesn't involve time:
![v^{2}= v_{0}^{2}+2a(x- x_{0})](https://tex.z-dn.net/?f=%20v%5E%7B2%7D%3D%20%20v_%7B0%7D%5E%7B2%7D%2B2a%28x-%20x_%7B0%7D%29%20%20%20%20)
In this equation x is the final position, which we take to be 0. Also the initial velocity Vo is zero. Thus the equation simplifies to
![v^{2}= 2a(- x_{0})](https://tex.z-dn.net/?f=%20v%5E%7B2%7D%3D%202a%28-%20x_%7B0%7D%29)
Putting in v=32m/s, a=-9.81m/s^2 gives
![32^{2}= 2(-9.81)(- x_{0}) ](https://tex.z-dn.net/?f=32%5E%7B2%7D%3D%202%28-9.81%29%28-%20x_%7B0%7D%29%0A)