The radius of curvature of the proton's path while in the field is 
× 
.
b) Let R = radius curvature of protons path. Then,
relation b/w B, R, and v is: -
×
Hence, the radius of curvature of the proton's path while in the field is × .
<h3>
What do you mean by Magnetic field?</h3>
The magnetic influence on moving electric charges, electric currents and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is travelling through a magnetic field. The magnetic field of a permanent magnet pulls on ferromagnetic substances like iron and attracts or repels other magnets. A magnetic field that varies with location will also exert a force on a variety of non-magnetic materials by changing the velocity of those particles' outer electrons. Electric currents, like those utilized in electromagnets, and electric fields that change in time produce magnetic fields that surround magnetized things.
To know more about Magnetic Field visit:
brainly.com/question/14848188
#SPJ4
The answer is C in this question.
Answer:
Explanation:
<u>Conservation of Momentum
</u>
The total momentum of a system of two particles is
Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now
The momentum is conserved if no external forces are acting on the system, thus
Let's put some numbers in the problem and say
120=120
It means that when the particles collide, the first mass returns at 6 m/s and the second continues in the same direction at 28 m/s
Evidence: Data gathered
Experiment: Looking through a telescope
Observations: Testing what happens
Reasoning: Thinking a problem through
I believe that these should be correct.
Hoping you pass!
Answer:
<h2>
3000 J</h2>
Option C is the correct option.
Explanation:
Given,
Force = 600 N
Distance = 5 meters
Work = ?
Now,
Work = Force 
distance
Calculate the product
Joule
Hope this helps...
Good luck on your assignment..
