(a) What is the average useful power output of a person who does 6.00×106 J of useful work in 8.00 h? (b) Working at this rate,
1 answer:
Answer:
(a) 208.33 J/s
(b) 141.12 s
Explanation:
(a) Energy is the rate of doing work.
Work is the product of power and time
Work = Power × Time
Power = Work/time
where work is in joule and time is in seconds
Given that work = 6 ×
Time = 8 h = 8 × 60 × 60
= 28800 seconds
Average useful power =
= 208.33 J/s
(b) Given
mass = 2000kg , using g = 9.8 m/s2
Force = 2000 × 9.8
= 19600N
Work = Force × Distance ( considering that Work done to lift his body can be omitted because it is not considered useful output here)
Work = 19600 × 1.5
= 29,400J
Time taken = Work/Power
=
= 141.12s
≈ 2 Minutes 21 Seconds
At the rate of 208.33J/s, it will take the person 141.12 seconds to lift 2000 kg of bricks 1.50 m to a platform.
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Solution :
a). At the break even units, the total contribution margin = fixed expenses
We know that : (Selling price - variable cost) x units sold = fixed expenses
i.e. (20-14)x = 225,000
6x = 225,000
x = 37,500
Therefore, the number of units sold, x = 37,500
So, the break even analysis = 37,500 x 20
= 750,000
b).
= 30%
The Breakeven sales =
= 750,000
c).
= 37.5%
d). Units needed :
units
Therefore, the sales required = 62,500 x 20
= 125,000