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aivan3 [116]
3 years ago
15

The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t

o within 0.1kmh. What is the potential error in her calculated kinetic energy, as a result of the imprecision in the measurement of her velocity?
Physics
1 answer:
jek_recluse [69]3 years ago
4 0

Answer:

e=3367.2J

%e=1.43%

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

v_{r}=10km/h=2.77m/s

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s

Knowing that we can calculate Kinetic energy for the real and experimental speed

E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J

E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J

Now, the potential error in her calculated kinetic energy is:

e=E_{r}-E_{e}=(234023-230656)J=3367.2J

%e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43%

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A plank rests on top of the axles of two identical wheels. Each wheel's outer radius is 0.25 meters and each wheel's axle has ra
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<u><em>The plank moves 0.2m from it's original position</em></u>

Explanation:

we can do this question from the constraints that ,

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<em>since the wheel is pure rolling or not slipping,</em>

<em>⇒v=wr</em>

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<em>v - speed of the wheel</em>

<em>w - angular speed of the wheel</em>

<em>r - radius of the wheel</em>

<em>since the wheel traverses 1 m let's say in time 't' ,</em>

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⇒w=\frac{v_{w}}{r} = \frac{1}{t*0.25}

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⇒v_{a}=w*r\\v_{a}=\frac{1}{t*0.25} *0.05\\v_{a}=\frac{1}{5t}

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T: Tangent
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Answer:

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