Answer:
A_resulting = 0.2 m
Explanation:
Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.
With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is
A_res = 2A
A_resultant = 2 .01
A_resulting = 0.2 m
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.
To learn more about the angular momentum, refer to the link;
brainly.com/question/15104254
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Answer:
2) c) give-way vessel
3) a) With one short blast
Explanation:
2) A vessel that is required to take early substantial action to ensure avoiding collision called Give way vessel
In overtaking, the vessel intending to overtake is the Give-Way Vessel the vessel that is going to be overtaken is the Stand-On Vessel
Therefore, the correct option is c) give-way vessel
3) When vessels use sound signals in a meeting head on situation both vessel are Give-Way vessels and both vessel pass the each other by turning to the starboard side therefore they intend to pass each other on their port side requiring one short blast
Therefore, the correct option is a) With one short blast.
Given data:
* The mass of the ball is 2 kg.
* The gravitational field strength at the surface of planet X is 5 N/kg.
Solution:
The weight of the ball on the planet X is,
where m is the mass of ball, a is the gravitational field strength,
Substituting the known values,
Thus, the weight of the ball on the surface of planet X is 10 N.
