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harina [27]
2 years ago
8

In a hard disk drive, a constant torque of 14 N⋅m is applied to the magnetic disk when the drive starts recording data. The magn

etic disk has a mass of 0.2 kg and radius 2m. Determine the magnitude of the angular acceleration of the disk? [please solve if you know]
Physics
1 answer:
marshall27 [118]2 years ago
7 0

The magnitude of the angular acceleration of the magnetic disk is determined as 35 rad/s².

<h3>Angular acceleration of the disk</h3>

The angular acceleration of the disk is determined from the principle of conservation of angular momentum as follows;

τ = Iα

where;

α is angular acceleration

I is moment of inertia of the disk

I = 0.5MR²

I = 0.5 x 0.2 x 2²

I = 0.4 kgm²

τ = Iα

α = τ/I

α = (14)/(0.4)

α = 35 rad/s²

Learn more about angular acceleration here: brainly.com/question/21278452

#SPJ1

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3 years ago
the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?
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Via the half-life equation:

A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}

Where the time elapse is 11,460 year and the half-life is 5,730 years.

A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.

6 0
2 years ago
An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu
ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, qE \times S + mg \times S = 0.5 \times mv^{2}

     1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

   (1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2&#10;}

         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

           = 592999 m/s

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Answer:

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     h = 1.02 m

This is the depth at which man can breathe

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