The amplitude of wave-c is 1 meter.
The speed of all of the waves is (12meters/2sec)= 6 m/s.
The period of wave-a is 1/2 second.
Answer:
0.187 m
Explanation:
We'll begin by calculating the acceleration of the ball. This can be obtained as follow:
Mass (m) = 0.450 Kg
Force (F) = 38 N
Acceleration (a) =?
F = m × a
38 = 0.450 × a
Divide both side by 0.450
a = 38 / 0.450
a = 84.44 m/s²
Finally, we shall determine the distance. This can be obtained as follow:
Initial velocity (u) = 2.20 m/s.
Final velocity (v) = 6 m/s
Acceleration (a) = 84.44 m/s²
Distance (s) =?
v² = u² + 2as
6² = 2.2² + (2 × 84.44 × s)
36 = 4.4 + 168.88s
Collect like terms
36 – 4.84 = 168.88s
31.52 = 168.88s
Divide both side by 168.88
s = 31.52 / 168.88
s = 0.187 m
Thus, the distance is 0.187 m
Answer:
B is the answer. Correct me if I'm wrong
The distance traveled by the hockey player is 0.025 m.
<h3>The principle of conservation of linear momentum;</h3>
- The principle of conservation of linear momentum states that, the total momentum of an isolated system is always conserved.
The final velocity of the hockey play is calculated by applying the principle of conservation of linear momentum;
The time taken for the puck to reach 15 m is calculated as follows;
The distance traveled by the hockey player at the calculated time is;
Learn more about conservation of linear momentum here: brainly.com/question/7538238
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