What kind of wave exists when a musician plays a note on a trombone?
1 answer:
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Answer:
The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South
Explanation:
The given collision parameters are;
The kind of collision experienced by the four velcro-lined air-hockey disk = Inelastic collision
The mass of the first disk, m₁ = 50.0 g
The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i
The mass of the second disk, m₂ = 60.0 g
The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j
The mass of the third disk, m₃ = 100.0 g
The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i
The mass of the fourth disk, m₄ = 40.0 g
The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j
Therefore, the total initial momentum of the four velcro-lined air-hockey disk, is given as follows;
= m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)
∴ = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j
∴ = -20·i + 130·j
By the law of conservation of linear momentum, we have;
= -20·i + 130·j
Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0
∴ m = 250.0 g
Let, "v" represent the final velocity of the four disks moving as one after the collision
We have;
= m × v = 250.0 × v = -20·i + 130·j
∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j
The final velocity of the four disks after collision, v = -0.08·i + 0.52·j
The magnitude of the final velocity, = √((-0.08)² + (0.52)²) ≈ 0.526
≈ 0.526 m/s
The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°
The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South
Answer:
The correct option is;
C. 40 N
Explanation:
Given that the rock is on the ground, we have;
The weight of the rock, = 40 N ↓
From Newton's third law of motion, we have that action and reaction are equal and opposite,
Therefore, given that the weight of the rock exerts a force of 40 N on the ground, and the rock is stable and stationery on the ground, then the ground is reacting to the weight of the rock with an equal force of 40 N, acting in opposite direction to the weight of the block to balance the weight of the rock and create equilibrium such that the net force acting on the rock is zero
∑F = = 0
Where;
= The reaction of the ground
We have;
+
= 40 +
= 0
= 0 - 40 = -40
= -40 N ↓
∴ = 40 N ↑.