The magnetic force experienced by the proton is given by
![F=qvB \sin \theta](https://tex.z-dn.net/?f=F%3DqvB%20%5Csin%20%5Ctheta)
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and
![\theta](https://tex.z-dn.net/?f=%5Ctheta)
the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so
![\sin \theta=1](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%3D1)
and we can ignore it in the formula.
For Netwon's second law, the force is also equal to the proton mass times its acceleration:
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
So we have
![ma=qvB](https://tex.z-dn.net/?f=ma%3DqvB)
from which we can find the magnitude of the field:
If
Sunny watched a video of her professor’s lecture on genes and environment the
night before class and her professor facilitated a discussion about the role of
nature and nurture with respect to development, Sunny is experiencing being
advanced in their lesson. Nurture plays an important part in this. The
environment poses an encouraging and stimulating atmosphere where Sunny might
enjoy learning about things.
Answer:
360 degrees is one full rotation starting at zero
Answer: 115.2kg
Explanation:
Net force = 265 N
Acceleration of bike & rider = 2.30m/s2 (The SI unit of acceleration is m/s2)
Mass of the bike and rider together = ?
Since force is the product of the mass of an object and the acceleration by which it moves, Force = Mass x Acceleration
265N = Mass x 2.30m/s2
Mass = (265N/2.30m/s2)
Mass = 115.2 kg
Thus, the Mass of the bike and rider together is 115.2kg
Answer:
![F_x=208.25\ N](https://tex.z-dn.net/?f=F_x%3D208.25%5C%20N)
Explanation:
Given that,
Mass of a crate is 22 kg
It moved up along the 15 degrees incline without tipping.
We need to find the corresponding magnitude of force P. The force P is acting in horizontal direction.
It means that the horizontal component of force is given by :
![F_x=F\cos\theta\\\\F_x=mg\cos\theta\\\\F_x=22\times 9.8\times \cos(15)\\\\F_x=208.25\ N](https://tex.z-dn.net/?f=F_x%3DF%5Ccos%5Ctheta%5C%5C%5C%5CF_x%3Dmg%5Ccos%5Ctheta%5C%5C%5C%5CF_x%3D22%5Ctimes%209.8%5Ctimes%20%5Ccos%2815%29%5C%5C%5C%5CF_x%3D208.25%5C%20N)
So, the horizontal component of force is 208.25 N.