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Svet_ta [14]
3 years ago
11

The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s. The centripetal acceleration of the

edge of the disc is ?m/s2.
Physics
2 answers:
olga_2 [115]3 years ago
8 0

Answer:

the answer is 84

Explanation:

Marat540 [252]3 years ago
5 0

Answer:

Centripetal acceleration = 83.77m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 0.13m

Velocity, v = 3.3m/s

To find centripetal acceleration;

Centripetal acceleration is given by the formula;

Acceleration, a = \frac {v^{2}}{r}

Substituting into the equation, we have;

Centripetal \; acceleration, a = \frac {3.3^{2}}{0.13}

Centripetal \; acceleration, a = \frac {10.89}{0.13}

<em>Centripetal acceleration = 83.77m/s²</em>

<em>Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s². </em>

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Answer:

\lambda=1.37\times 10^{-11}\ m

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Give that,

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Using the conservation of energy,

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Put all the values,

\lambda=\sqrt{\dfrac{(6.63\times 10^{-34})^2}{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 8\times 10^3}}\\\\\lambda=1.37\times 10^{-11}\ m

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3 years ago
What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?
kvv77 [185]

Answer:

W = 78.48N\\W =0.0784kN\\W = 8kgf\\

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Explanation:

if

m=8kg=3.6lb\\

and g=9.81 m/s2=32.16 ft/s2

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we can just replace de mass and gravity and we have

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<em><u>because if any metals are there in the flour it would attract it...</u></em>

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