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:
x = 50 N
Explanation:
Given that we have a net force, a mass, and acceleration, we can use the fundamental formula for force found in newton's second law which is F = m × a.
Given a mass of 150 kg, and an acceleration 3.0m/s². We can substitute these two values in our formula to calculate the magnitude of these forces or it's net force to identify the unknown force acting on our known force for this situation to work.
_______
F (Net force) = F2 (Second force which we are given) - F1 (First force) = m × a
m (mass which we are given) = 150 kg
a (acceleration which we are given) = 3.0m/s
________
So F = m × a → F2 - F1 = m × a →
500 - F1 = 150 × 3.0 → 500 - F1 = 450 →
-F1 = -50 → F1 = 50
Answer:
Moment of inertia of the solid sphere:
I
s
=
2
5
M
R
2
.
.
.
.
.
.
.
.
.
.
.
(
1
)
Is=25MR2...........(1)
Here, the mass of the sphere is
M
M
Answer:
A decrease in the distance between the earth and the moon.
Answer:
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