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Anna35 [415]
3 years ago
9

(10%) Problem 5: The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass

of the Earth: 5.97 × 1024 kg (assume a uniform mass distribution) Radius of the Earth: 6371 km Distance of Earth from Sun: 149,600,000 km
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0

Incomplete Question.The Complete question is

The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants.  Mass of the Earth: 5.97 × 10^24  kg (assume a uniform mass distribution)  Radius of the Earth: 6371 km  Distance of Earth from Sun: 149,600,000 km

(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.

(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?  

Answer:

(i) KE= 2.56e29 J

(ii) KE= 2.65e33 J

Explanation:

i) Treating the Earth as a solid sphere, its moment of inertia about its axis is

I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²

I = 9.69e37 kg·m²

About its axis,

ω = 2π rads/day * 1day/24h * 1h/3600s

ω= 7.27e-5 rad/s,

so its rotational kinetic energy

KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²

KE= 2.56e29 J

(ii) About the sun,

I = mR²

I= 5.97e24kg * (1.496e11m)²

I= 1.336e47 kg·m²

and the angular velocity

ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s

ω= 1.99e-7 rad/s

so  

KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²

KE= 2.65e33 J

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The work function for magnesium is 3.70 ev. what is its cutoff frequency?
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<h3>What is cutoff frequency?</h3>

The work function is related to the frequency as

W0 = h x fo

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Given is the work function for magnesium is  3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

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brainly.com/question/14378802

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1 year ago
astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating o
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Answer:

The right answer is:

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Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ T=2 \pi\sqrt{\frac{m}{k} }

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⇒ m=\frac{k T^2}{4 \pi^2}

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(b)

The speed of the string will be:

⇒  \frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2

then,

⇒             v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }

On substituting the values, we get

⇒                =\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }

⇒                =\sqrt{\frac{280(0.16-0.04)}{63.83} }

⇒                =\sqrt{\frac{280\times 0.12}{63.83} }

⇒                =0.725 \ m/s

4 0
3 years ago
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