Physics
25.02.2023 19:00
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You find a micrometer (a tool used to measure objects to the nearest 0.01 mm) that

You find a micrometer (a tool used to measure objects to the nearest 0.01 mm) that has been badly bent. how would it compare to a new, high-quality meterstick in terms of its precision? its accuracy?
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benwill0702
benwill0702
4,9(60 marks)

The error which is find in the damaged micrometer would be less than as compared to the width of the hash marks on the stick.

Explanation:

If the micrometer is still in the working condition, it will be hard for the observer to find its accuracy or precision, which could be matched with the meter stick. Its accuracy can be compromised in comparison with the undamaged micrometer, but it would be more chances of highly precise.

The error which is find in the damaged micrometer would be less than as compared to the width of the hash marks on the stick.

Blossom824
Blossom824
4,5(91 marks)

Part a)

v_{cm} = 2.98 \times 10^4 m/s

Part b)

K_{trans} = 2.68 \times 10^{33} J

Part c)

\omega = 7.27 \times 10^{-5} rad/s

Part d)

KE_{rot} = 2.6 \times 10^{29} J

Part e)

KE_{tot} = 2.68 \times 10^{33} J

Explanation:

Time period of Earth about Sun is 1 Year

so it is

T = 1 year = 3.15 \times 10^7 s

now we know that angular speed of the Earth about Sun is given as

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{3.15 \times 10^7}

now speed of center of Earth is given as

v_{cm} = r\omega

r = 1.5 \times 10^{11} m

v_{cm} = (1.5 \times 10^{11})(\frac{2\pi}{3.15 \times 10^7})

v_{cm} = 2.98 \times 10^4 m/s

Part b)

now transnational kinetic energy of center of Earth is given as

K_{trans} = \frac{1}{2}mv^2

K_{trans} = \frac{1}{2}(6 \times 10^{24})(2.98 \times 10^4)^2

K_{trans} = 2.68 \times 10^{33} J

Part c)

Angular speed of Earth about its own axis is given as

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{24 \times 3600}

\omega = 7.27 \times 10^{-5} rad/s

Part d)

Now moment of inertia of Earth about its own axis

I = \frac{2}{5}mR^2

I = \frac{2}{5}(6 \times 10^{24})(6.4 \times 10^6)^2

I = 9.83 \times 10^{37} kg m^2

now rotational energy is given as

KE_{rot} = \frac{1}{2}I\omega^2

KE_{rot} = \frac{1}{2}(9.83 \times 10^{37})(7.27 \times 10^{-5})^2

KE_{rot} = 2.6 \times 10^{29} J

Part e)

Now total kinetic energy is given as

KE_{tot} = KE_{trans} + KE_{rot}

KE_{tot} = 2.68 \times 10^{33} + 2.6 \times 10^{29}

KE_{tot} = 2.68 \times 10^{33} J

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