C. Newtons third law of motion
Because eventually, the frictional forces will slow you to a halt. Newton's Third Law of Motion For every action there is an equal and opposite reaction. When they push off against the ice, or "stroke" with their skates, they are applying a force down and back against the ground.
Hope this helps!
Answer:
![9.6\cdot 10^{-11}N](https://tex.z-dn.net/?f=9.6%5Ccdot%2010%5E%7B-11%7DN)
Explanation:
The force experienced by a charged particle immersed in an electric field is
![F=qE](https://tex.z-dn.net/?f=F%3DqE)
where
q is the charge of the particle
E is the strength of the electric field
For a uniform field, we can write
![E=\frac{\Delta V}{d}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B%5CDelta%20V%7D%7Bd%7D)
where
is the potential difference
d is the distance travelled
So we can rewrite the force as
![F=\frac{q\Delta V}{d}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7Bq%5CDelta%20V%7D%7Bd%7D)
In this problem:
is the charge of the electron
is the potential difference
d = 0.01 m is the distance covered
Substituting, we find the force:
![F=\frac{(1.6\cdot 10^{-19})(6\cdot 10^6)}{0.01}=9.6\cdot 10^{-11}N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B%281.6%5Ccdot%2010%5E%7B-19%7D%29%286%5Ccdot%2010%5E6%29%7D%7B0.01%7D%3D9.6%5Ccdot%2010%5E%7B-11%7DN)
Answer:
Mass: Choices 1, 2, 4, 6
Weight: Choices 3, 5, 7, 8
Hope this helped! Please mark Brainliest if correct.
Answer:
Magnetic field at the center of the loop ![B=5.89\times 10^{-5}\ T.](https://tex.z-dn.net/?f=B%3D5.89%5Ctimes%2010%5E%7B-5%7D%5C%20T.)
Explanation:
It is given that total length of wire is 2 m and number of circular loop is 5 turns.
Therefore ,
![5\times ( 2\pi r)=2 \ m .\\\\r=\dfrac{1}{5 \pi}=0.064\ m.](https://tex.z-dn.net/?f=5%5Ctimes%20%28%202%5Cpi%20r%29%3D2%20%5C%20m%20.%5C%5C%5C%5Cr%3D%5Cdfrac%7B1%7D%7B5%20%5Cpi%7D%3D0.064%5C%20m.)
We know , magnetic field at the center of loop is given by :
![B=N\dfrac{\mu_o i}{2r}](https://tex.z-dn.net/?f=B%3DN%5Cdfrac%7B%5Cmu_o%20i%7D%7B2r%7D)
Putting all values in above equation we get :
![B=5\times \dfrac{4\pi\times 10^{-7}\times 1.2}{2\times 0.064}\\\\B=5.89\times 10^{-5}\ T.](https://tex.z-dn.net/?f=B%3D5%5Ctimes%20%5Cdfrac%7B4%5Cpi%5Ctimes%2010%5E%7B-7%7D%5Ctimes%201.2%7D%7B2%5Ctimes%200.064%7D%5C%5C%5C%5CB%3D5.89%5Ctimes%2010%5E%7B-5%7D%5C%20T.)
Hence , this is the required solution.
Using a graphical tool to obtain the form of the wave we can see that
It is periodic in X with period = 5 cm
It is periodic in t with period = 1/125 s = 8 ms