Question: 1)All of the following are 2 component models EXCEPT... A. Hydrostatic weighing (underwater weighing) B. Skinfold thickness C. Air-displacement plethysmography (bod-pod) D. Dual-energy x-ray absorptiometry
1)All of the following are 2 component models EXCEPT...
| A. | Hydrostatic weighing (underwater weighing) | |
| B. | Skinfold thickness | |
| C. | Air-displacement plethysmography (bod-pod) | |
| D. | Dual-energy x-ray absorptiometry (DEXA) |
2)Which method of densitometry utilizes Archimedes' principle to determine body density?
| A. | Skinfold thickness | |
| B. | Bioelectrical Impedance Analysis (BIA) | |
| C. | Air-displacement plethysmography (bod-pod) | |
| D. | Hydrostatic weighing (underwater weighing) |
3)
Which of the following could best (most accurately) measure vertical power output?
| A. | Jump test using a wall | |
| B. | Jump test on a force plate | |
| C. | Jump test using a Vertec | |
| D. | Jump test on a switch mat |
4)
The next day at your facility, a sprinter comes in and wants to test their anaerobic power. You are going to be sport specific, so you do a sprint test rather than a WAnT. The sprinter is 177cm tall and 72kg. The athlete’s resting HR is 58BPM, and resting blood pressure is 118/76. The sprinter runs at top speed over 100m. The athlete completes the distance in 14 seconds.
You want to calculate the average horizontal power (in kgm/s) for the 100m. This requires the mass moved to be multiplied by the speed at which it was moving. What would be this athlete's horizontal power?
| A. | ~557kgm/s | |
| B. | ~514kgm/s | |
| C. | ~613kgm/s | |
| D. | ~479kgm/s |
5)
Use the information from questions 61-64. Power output for the wingate test can be calculated by multiplying the kg resistance x 9.80665 x number of revolutions x distance of the flywheel (typically 6m) ÷ the duration of cycling. This can be difficult, so algebraically, this could be reduced to kg x rpm x 58.8399 ÷ the duration of cycling. If you were to measure power output for 5 seconds, where the cyclist performed 11 revolutions, what would the power output be over this period? (hint: 58.8399 ÷ 5 = 11.768)
| A. | ~789W | |
| B. | ~737W | |
| C. | ~854W | |
| D. | ~822W |
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