Answer:
2240.92365 m/s
Explanation:
= Mass of electron = 
= Speed of electron = 
= Neutrino has a momentum = 
M = total mass = 
In the x axis as the momentum is conserved

In the y axis

The resultant velocity is

The recoil speed of the nucleus is 2240.92365 m/s
Answer:
y = 10.2 m
Explanation:
It is given that,
Charge, 
It is placed at a distance of 9 cm at x axis
Charge, 
It is placed at a distance of 16 cm at x axis
We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

Here,

So,

Squaring both sides,

So, at a distance of 10.2 m on the y axis the electric potential equals 0.
Answer:
The initial velocity was U=22.14m/s
Explanation:
Step one :
Applying the third equation of motion
v² = u²+ 2as
Where v= Final velocity
U =initial velocity
a= acceleration due to gravity
S= distance or displacement
Step two :
V= 0
a= 9.81m/s²
S=25m
U=?
Step three :
Substituting into the equation we have
0²=U²+2*9.81*25
0=U²+490.5
U²=-490.5
U=√490.5
U=22.14m/s
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Is there a multiple choice?