A 76-kg man standing on a scale in an elevator notes that as the elevator rises, the scale reads 840 N. What is the acceleration
1 answer:
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Answer:
u= 20.09 m/s
Explanation:
Given that
m = 0.02 kg
M= 2 kg
h= 0.2 m
Lets take initial speed of bullet = u m/s
The final speed of the system will be zero.
From energy conservation
1/2 m u²+ 0 = 0+ (m+M) g h
m u²=2 (m+M) g h
By putting the values
0.02 x u² = 2 (0.02+2) x 10 x 0.2 ( take g=10 m/s²)
u= 20.09 m/s
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}