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Ulleksa [173]
2 years ago
14

Using the formula for kinetic energy of a moving particle k=12mv2, find the kinetic energy ka of particle a and the kinetic ener

gy kb of particle
b. remember that both particles rotate about the y axis.
Physics
1 answer:
Flauer [41]2 years ago
5 0
<span>Answer: KE = (11/2)mω²r², particle B must have mass of 2m, while A has mass m. Then the moment of inertia of the system is I = Σ md² = m*(3r)² + 2m*r² = 11mr² and then KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2 So I'll proceed under that assumption. For particle A, translational KEa = ½mv² but v = ω*d = ω*3r, so KEa = ½m(3ωr)² = (9/2)mω²r² For particld B, translational KEb = ½(2m)v² but v = ω*r, so KEb = ½(2m)ω²r² so total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2 which is equal to our rotational KE.</span>
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Answer:

B. Solar energy

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Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Which student’s measureme
Lisa [10]

On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2 .</u>

<u>Explanation:</u>

Here, we have Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Let's calculate which student’s measurement has the largest percent error :

<u>A. Student 4: 9.61 m/s2 </u>

Percentage of error  = \frac{9.78-9.61}{9.78} (100) = 1.738% %.

<u>B. Student 3: 9.88 m/s2 </u>

Percentage of error  = \frac{9.88-9.78}{9.78} (100) = 1.022% %.

<u>C. Student 2: 9.79 m/s2 </u>

Percentage of error  = \frac{9.79-9.78}{9.78} (100) = 0.1022%% .

<u>D. Student 1: 9.78 m/s2</u>

Percentage of error  = \frac{9.78-9.61}{9.78} (100) = 0%% .

On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2 .</u>

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