C because when the part gets out of the probe it would no longer stay contacted
Answer:C
Explanation:
Power=280watts=280/1000 kilowatts
Power=0.28 kilowatts
Device use 0.28 kilowatts in 1 day
1day=24hours
Device use 0.28 kilowatts in 24hours
30days=30 x 24=720 hours
For 720 hours=(0.28x720) ➗ 24
For 720 hours=201.6 ➗ 24
For 720 hours=8.4
The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
<em>W</em> = ∆<em>K</em> = <em>K</em>
where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
<em>W</em> = 1/2 <em>mv</em> ²
where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get
<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s
Carbohydrates, in cellular respiration.
