The magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.
<h3>What is angular momentum.?</h3>
The rotational analog of linear momentum is angular momentum also known as moment of momentum or rotational momentum.
It is significant in physics because it is a conserved quantity. the total angular momentum of a closed system remains constant. Both the direction and magnitude of angular momentum are conserved.
The magnitude of the angular momentum of the two-satellite system is best represented as;
L=∑mvr
L=m₁v₁r₁-m₂v₂r₂
Hence, the magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.
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Answer:
A. 50 m/s
Explanation:
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 4 s
Find: v
v = at + v₀
v = (10 m/s²) (4 s) + 0 m/s
v = 40 m/s
In the x direction, the velocity is constant at 30 m/s.
The overall speed is:
v² = (30 m/s)² + (40 m/s)²
v = 50 m/s
Kepler's second law of planetary motion<span> describes the speed of a </span>planet<span> traveling in an elliptical orbit around the sun. It states that a line between the sun and the </span>planetsweeps equal areas in equal times. Thus, the speed of theplanet<span> increases as it nears the sun and decreases as it recedes from the sun.</span>
By Boyle's law:
P₁V₁ = P₂V₂
70*8 = P<span>₂*4
</span>P<span>₂*4 = 70*8
</span>
P<span>₂ = 70*8/4 = 140
</span>
P<span>₂ = 140 kiloPascals.</span>
Answer:
α(0) = 0 rad/s²
α(5) = 15 rad/s²
Explanation:
The angular velocity of the flywheel is given as follows:
w(t) = A + B t²
where, A and B are constants.
Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:
Angular Acceleration = α (t) = dw/dt
α(t) = (d/dt)(A + B t²)
α(t) = 2 B t
where,
B = 1.5
<u>AT t = 0 s</u>
α(0) = 2(1.5)(0)
<u>α(0) = 0 rad/s²</u>
<u></u>
<u>AT t = 5 s</u>
α(5) = 2(1.5)(5)
<u>α(5) = 15 rad/s²</u>
