Answer:
4347.8 m/s
Explanation:
It is given that the radius of the circular path traversed by proton and electron is same. Also, we know that magnitude of charge on an electron and proton is same. Magnetic field strength is same for both.

Take the ratio:

The momentum of an object is given by:
p = mv
m is the object's mass and v is its velocity.
Given values:
m = 15kg
v = 25m/s
Plug in the values and solve for p:
p = 15*25
p = 375kg×m/s
Answer:
The intensity of laser 2 is 4 times of the intensity of laser 1.
Explanation:
The intensity in terms of electric field is given by :

E is electric field
It means, 
In this problem, lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of laser 2.
Let E is electric field in the beam of laser 1 and E' is the electric field in the beam of laser 2. So,

We have,
E'=2E
So,

So, the intensity of laser 2 is 4 times of the intensity of laser 1.