Answer:All professors planned and thought about data before graph construction. When reflecting on their graphs, professors and graduate students focused on the ... feel more inclusive in the learning process and gain better science process skills, ... We organized data in a table instead of a paragraph with numbers
Explanation:
hope that helps
The formula for frequency is f = 1/T where f is frequency and T is period in seconds.
You have you period which is 0.008s and that is all you will need to solve or frequency in a wave:
f = 1/2
f = 1/0.008s
f = 125Hz
Protons are positive, and neutrons are negative, electrons are neutral. I’m not sure about the rest but I hope that helps for now
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