Answer:
r = 3.787 10¹¹ m
Explanation:
We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration
F = ma
G m M / r² = m a
The centripetal acceleration is given by
a = v² / r
For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship
v = d / t
The distance traveled Esla orbits, in a circle the distance is
d = 2 π r
Time in time to complete the orbit, called period
v = 2π r / T
Let's replace
G m M / r² = m a
G M / r² = (2π r / T)² / r
G M / r² = 4π² r / T²
G M T² = 4π² r3
r = ∛ (G M T² / 4π²)
Let's reduce the magnitudes to the SI system
T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)
T = 1.03 10⁸ s
Let's calculate
r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]
r = ∛ (21.44 10³⁵ / 39.478)
r = ∛(0.0543087 10 36)
r = 0.3787 10¹² m
r = 3.787 10¹¹ m
Answer:
I would love to be a gardener :)
Explanation:
Plants are kool!!!
Answer:
The Force between the two charges is an attractive force of 16,000N
Explanation:
Expression for the electric force between the two charges is given by
F = (k*q1*q2) / r^2
Here, k = constant = 9 x 10^9 N*m^2 / C^2
q1 = - 2.0x10^-4C
q2 = + 8.0x10^-4C
r = 0.30 m
Substitute the given values in the above expression -
One charge is + and the other is a -, therefore the net force is an attractive force (opposites atract)
The attraction force is:
F= 9.0x10^9 * 2.0x10^-4 *8.0x10^-4 N/ 0.30^2
F= 16,000N
