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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Answer: The coefficient of kinetic friction is μ = 0.6
Explanation:
For an object of mass M, the weight is:
W = M*g
where g is the gravitational acceleration: g = 9.8m/s^2
And the friction force between this object and the surface can be written as:
F = W*μ
where μ is the coefficient of friction (kinetic if the object is moving, and static if the object is not moving, usually the static coefficient is larger)
In this case, the weight is:
W = 20N
And the friction force is:
F = 12N
Replacing these values in the equation for the friction force we get:
12N = 20N*μ
(12N/20N) = μ = 0.6
The coefficient of kinetic friction is μ = 0.6
Answer:
Perform a simple test of the material in the pan to assess whether it is real gold. Raw gold appears brassy yellow and bright. If you think it is gold, place your hand between it and the sun to create shade over the gold. If it still appears bright in the pan, chances are that it is real gold.
Explanation: