The key principle is that crank length, just like frame size, should be proportional to the rider height and then modified to what fits the individual. There are 4 charts, two for the upright position and two for the aero position, depending upon how you race.
The magnitude and direction (inward or outward) of the net flux through the cell boundary is - 0.887 wb.m².
<h3>
What is flux?</h3>
Flux describes any effect that appears to pass or travel through a surface or substance.
The magnitude and direction (inward or outward) of the net flux through the cell boundary is calculated as follows;
Ф = Q/ε
where;
- Q is net charge
- ε is permittivity of free space
Φ = (-7.85 x 10⁻¹²)/(8.85 x 10⁻¹²)
Φ = - 0.887 wb.m²
Learn more about flux here: brainly.com/question/10736183
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V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s
v=d/t
d=vt
d=(11111.1mm/s)(5s)
d=55555mm
d=5.56x10^4mm
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
