A plane is going at a speed of 300 km/h at 63 W of N. The wind hits the plane at a direction of 65 km/h at 52 S of E. What is th
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Answer:
(a)2.7 m/s
(b) 5.52 m/s
Explanation:
The total of the system would be conserved as no external force is acting on it.
Initial momentum = final momentum
⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)
⇒ 730 ×v = (4054.9 - 2081.2) =1973.7
⇒v=2.7 m/s
Thus, the resulting speed of the block is 2.7 m/s.
(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.
Thus, the speed of the bullet-block center of mass is 5.52 m/s.
The maximum speed of the object under simple harmonic motion is 0.786 m/s.
The given parameters:
- Position of the particle, y = 0.5m sin(πt/2)
y = A sin(ωt + Ф)
where;
- A is the amplitude = 0.5 m
- ω is the angular speed = π/2
The maximum speed of the object is calculated as follows;
Thus, the maximum speed of the object under simple harmonic motion is 0.786 m/s.
Learn more about simple harmonic motion here: brainly.com/question/17315536