Answer:
The potential for r > rb is equal to zero.
Explanation:
For r > rb, the potential is:
![V=\frac{Kq}{r}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BKq%7D%7Br%7D)
Then, the net potential is:
![V_{(r)} =\frac{K(+\epsilon )}{r} +\frac{K(-\epsilon )}{r}](https://tex.z-dn.net/?f=V_%7B%28r%29%7D%20%3D%5Cfrac%7BK%28%2B%5Cepsilon%20%29%7D%7Br%7D%20%2B%5Cfrac%7BK%28-%5Cepsilon%20%29%7D%7Br%7D)
![K=\frac{1}{4\pi \epsilon _{o} }](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon%20_%7Bo%7D%20%20%7D)
![V_{(r)} =\frac{K(+\epsilon )}{r} -\frac{K(\epsilon )}{r}\\V_{(r)}=0](https://tex.z-dn.net/?f=V_%7B%28r%29%7D%20%3D%5Cfrac%7BK%28%2B%5Cepsilon%20%29%7D%7Br%7D%20-%5Cfrac%7BK%28%5Cepsilon%20%29%7D%7Br%7D%5C%5CV_%7B%28r%29%7D%3D0)
Answer:
The only one that makes sense is A.
Explanation:
Please help me by marking me brainliest. I'm one away :)
A - the amount of time needed to travel the distance
The formula for speed is distance/time and the unit is m/s. Therefore you divide the distance travelled by the time it took to travel it.
Answer:
Explanation:
Impulse results in change of momentum
FΔt = mΔv
F = 0.030(50.0 - 0) / 0.010 = 150 N
The vertical portion of that force adds to his weight on the ice
Fmax = 60.00(9.81) + 150sin30 = 663.6 N
As this is above the ice strength, He takes a cold dip.
Answer:14 m
Explanation:
Given
Vertical jump make by the dolphin is given by ![h=7\ m](https://tex.z-dn.net/?f=h%3D7%5C%20m)
Suppose the dolphin jump with an initial velocity of ![u](https://tex.z-dn.net/?f=u)
so u is given by ![u^2=2\cdot g\cdot h](https://tex.z-dn.net/?f=u%5E2%3D2%5Ccdot%20g%5Ccdot%20h)
If dolphin launches at an angle
then maximum horizontal range is given by
assuming the of Dolphin to be Projectile so range is given by
![R=\frac{u^2\sin 2\theta }{g}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7Bu%5E2%5Csin%202%5Ctheta%20%7D%7Bg%7D)
substitute the value of ![u^2](https://tex.z-dn.net/?f=u%5E2)
![R=\frac{2\times 9.8\times 7\sin 2\theta }{9.8}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B2%5Ctimes%209.8%5Ctimes%207%5Csin%202%5Ctheta%20%7D%7B9.8%7D)
![R=2h\sin 2\theta](https://tex.z-dn.net/?f=R%3D2h%5Csin%202%5Ctheta%20)
Range will be maximum for ![\theta =45^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D45%5E%7B%5Ccirc%7D)
thus ![R_{max}=2\times 7\times 1=14\ m](https://tex.z-dn.net/?f=R_%7Bmax%7D%3D2%5Ctimes%207%5Ctimes%201%3D14%5C%20m)