After the warm-up the participant performs the WAnT. You don't have equipment to automatically track power output, therefore, you count revolutions for each 5 second period of the test. The results are as follows: Time(sec) 5sec_rpm cumulative
0-5 10 10
5-10 7 17
10-15 9 26
15-20 7 33
20-25 7 40
25-30 6 46

The power output can now be computed for each 5 second period using the following equation:

kg of resistance x number of revolutions x 11.765= power (watts)

Required:
What is the power output for each 5 second period?

Missing detail ( weight of resistance = 6.69 Kg )

answer :

Step-by-step explanation:

Applying the equation for computing the power output

= Kg of resistance * number of revolutions * 11.765   ( 1 )

Calculate the power output for each 5 second period

weight ( Kg of resistance ) = 6.69 Kg

Equation 1 becomes :

6.69 * number of revolutions * 11.765   ( 2 )

Next :Apply equation (2) to every 5 seconds interval as represented in the table

For Time 0 - 5  

Number of revolutions ( rpm ) = 10

power output = 6.69 * 10 * 11.765  = 787.0785 watts

cumulative rpm = 10 + 0

Cumulative Power = power output + previous power output = 787. 0785watt

For Time  ( 5 - 10 )

Number of revolutions ( rpm ) = 7

Power output = 6.69 * 7 * 11.765 = 550.95 watts

Cumulative rpm = rpm + previous rpm = 7 + 10 = 17

Cumulative power = power output + previous power output

                               = 550.95 + 787.0785  = 1338.0285 watts

Power output for ( 10 - 15 ) = 6.69 * 9 * 11.765  = 708.37 watts

Power output for ( 15 - 20 ) = 6.69 * 7 * 11.765  = 550.95 watts

Power output for ( 20 - 25 ) = 6.69 * 7 * 11.765 = 550.95 watts

Power output for ( 25 - 30 ) = 6.69 * 6 * 11.765 = 472.25 watts

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

the 3rd on is correct

step-by-step explanation: