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:
The focal length fe of the eyepiece is <em>2.86 cm</em>
Explanation:
Since we are given the telescope's magnification and the length of the tube, we can use the expressions
<em>M = f_o/fe (1)</em> and
<em>l = f_o + fe (2)</em>
where
- M is the telescope's magnification
- l is the length of the tube
- fe is the focal length of the eye-piece
Rearranging equation (2) to make f_o the subject of the formula, we get
<em>f_o = l - fe</em>
Substituting the above equation into equation (1) we get
<em>M = (l - fe)/fe ⇒ fe = l/(M +1)</em>
<em> ⇒ fe = 60/(20 + 1)</em>
⇒ <em>fe = 2.86 cm</em>
Answer:
2. displacement is maximum
Explanation:
This makes sense as the spring is stretched the farthest at this point
F = kx
The largest force makes the largest acceleration.
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Answer:
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Explanation:
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