Answer:
The expresion for the flux through the disk is:
Ф = E·πR^2·cos(Θ).
Explanation:
Let's sat the electric field has direction e and the normal to the disk has direction n (bold means vector quantities). So we have:
E=E·e (where E is the magnitud of the electric flied)
A=A·n
The flux for an uniform electric field and a flat surface is:
Ф=E×A
⇒ Ф = E·A·e×n = E·A·cos(angle(e,n)) = E·A·cos(Θ)
Since in this case the area is for a disk of radius R, 
So, Ф = E·πR^2·cos(Θ)
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
Average acceleration = (change in speed) / (time for the change) .
Change in speed = (ending speed) - (beginning speed)
= (9.89 miles/hour) - (2.35 yards/second) = 26,839.2 ft/hr
Acceleration = (26,839.2 ft/hr) / (4.67 days) = 2,873.58 inch/hour²
