A weight lifter lifts a dumbbell a certain height in 2.0 s, while a competitor does the same workin 1.0 s. Compared to the power
1 answer:
Answer:
a. one-half as great
Explanation:
The power developed by the first lifter is one-half as great as that of the second person.
Power is defined as the rate at which work is done;
Power =
Since the two lifters do the same work at different time, let us estimate their power;
P₁ = P₂ =
We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.
Therefore,
The power of the first weight lifter is one-half the second lifter.
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The question is missing a diagram of the ray reflection. I attached a diagram which comes from a similar question in the answer section. The full question should be as follows:
Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point d = 10.0cmfrom their point of intersection, as shown in the figure. For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long) after reflecting from the first mirror?
Answer:
34.6°
Explanation:
To strike the midpoint of the second mirror, the ray light will have to travel half of the distance vertically
i.e. 29/2 = 14.5
We can solve this through trigonometry.
Let the angle between the ray and the vertical plane mirror is known as α
tan α = 10/14.5
α = = 34.6°
The angle of incidence is the angle between the ray and the normal line of the mirror.
Let angle of incidence of first mirror be β
β = α = 34.6
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
where q is the charge of the proton,
, with 
being the elementary charge, and 
and 
are the initial and final voltage.
Substituting, we get (in electronvolts):
and in Joule:
where q is the charge of the proton,
Substituting, we get (in electronvolts):
and in Joule: