There are two types of electric charges; positive and negative
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Answer:
a) fem = - 2.1514 10⁻⁴ V, b) I = - 64.0 10⁻³ A, c) P = 1.38 10⁻⁶ W
Explanation:
This exercise is about Faraday's law
fem = ![- \frac{ d \Phi_B}{dt}](https://tex.z-dn.net/?f=-%20%5Cfrac%7B%20d%20%5CPhi_B%7D%7Bdt%7D)
where the magnetic flux is
Ф = B x A
the bold are vectors
A = π r²
we assume that the angle between the magnetic field and the normal to the area is zero
fem = - B π 2r dr/dt = - 2π B r v
linear and angular velocity are related
v = w r
w = 2π f
v = 2π f r
we substitute
fem = - 2π B r (2π f r)
fem = -4π² B f r²
For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T
we reduce the magnitudes to the SI system
f = 2 rev / s (2π rad / 1 rev) = 4π Hz
we calculate
fem = - 4π² 428 10⁻⁶ 4π 0.10²
fem = - 16π³ 428 10⁻⁶ 0.010
fem = - 2.1514 10⁻⁴ V
for the current let's use Ohm's law
V = I R
I = V / R
I = -2.1514 10⁻⁴ / 0.00336
I = - 64.0 10⁻³ A
Electric power is
P = V I
P = 2.1514 10⁻⁴ 64.0 10⁻³
P = 1.38 10⁻⁶ W
Answer: Energy from the core travels by radiation through the radiative
zone, then by convection through the convection zone.
Explanation:
The average adult in the us spends 24 hours watching televistion each week
Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:
![\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}](https://tex.z-dn.net/?f=%20%5Cfrac%7BT_%7B1%7D%5E%7B2%7D%7D%7BT_%7B2%7D%5E%7B2%7D%7D%20%3D%20%5Cfrac%7Br_%7B1%7D%5E%7B3%7D%7D%7Br_%7B2%7D%5E%7B3%7D%7D%20)
<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!