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Answer:
Range, 
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}
Answer:
v = 0.489 m/s
Explanation:
It is given that,
Mass of a box, m = 1.5 kg
The compression in the spring, x = 6.5 cm = 0.065 m
Let the spring constant of the spring is 85 N/m
We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.


So, the speed of the box is 0.489 m/s.
Answer:
The direction of the magnetic force on a moving charge is perpendicular to the plane formed by v and B and follows right hand rule–1 (RHR-1)
Explanation:
hope this helps
Answer:
Θ
Θ
Θ = 
Explanation:
Applying the law of conservation of momentum, we have:
Δ

Θ (Equation 1)
Δ

Θ (Equation 2)
From Equation 1:
Θ
From Equation 2:
sinΘ = 

Replacing Equation 3 in Equation 4:


Θ (Equation 5)
And we found Θ from the Equation 5:
tanΘ=
Θ=