The work output of a machine divided by the work input is the ____ of the machine.
1 answer:
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Before answering this question, first we have to understand the effect of ratio of surface area to volume on the rate of diffusion.
The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-
V∝ R [ where v is the rate of diffusion and r is the ratio of surface area to volume]
As per the question,the ratio of surface area to volume for a sphere is given
The surface area to volume ratio for right circular cylinder is given
Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.
Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.
To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.
In the case of work, we know that it is defined by,
Where,
F= Force
d = Distance
The distance in this case is a composition between number of steps and the height. Then,
, for h as the height of each step and N number of steps.
On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)
Where,
Final velocity
Initial Velocity
a = Acceleration
Displacement
PART A) For the particular case of work we know then that,
Therefore the Work to do that activity is 4.41kJ
PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,
Here,
3 steps in one second
Replacing,
Re-arrange for a,
At this point we can calculate the time, which is,
With time and work we can finally calculate the power
P = \frac{W}{t} = \frac{4.41}{20}
P = 0.2205kW