Sinusoidal water waves are generated in a large ripple tank. The waves travel at 20 cm/s and their adjacent crests are 5.0 cm ap
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Answer:
Explanation:
Given
launch velocity
Launch angle
Range of Projectile
Horizontal velocity remain same and only vertical velocity changes.
Initially vertical velocity is in upward direction but as soon as it reaches the ground its direction change but magnitude remain same.
Refer to the diagram shown below.
F = 2000 N, the force exerted on the cannonball
L = 2.41 m, the length of the barrel
V₀ = 83 m/s, the launch velocity
θ = 35°, the launch angle
Let m = the mass of the cannonball.
Let a = the acceleration of the ball when fired.
The net force acting on the ball is
F - mg sinθ = 2000 - 9.8*m*sin(35°) = 2000 - 5.621*m N
Then, from Newton's Law of motion,
F = ma
2000 - 5.621*m = m*a (1)
The launch velocity is V₀ = 8.3 m/s, therefore
V₀² = 2*a*L
(83 m/s)² = 2*(a m/s²)*(2.41 m)
6889 = 4.82*a
a = 6889/4.82 = 1429.25 m/s² (2)
Insert (2) into (1)
2000 - 5.621*m = 1429.25*m
2000 = 1434.871*m
m = 1.394 kg
Answer: 1.394 kg
F = 2000 N, the force exerted on the cannonball
L = 2.41 m, the length of the barrel
V₀ = 83 m/s, the launch velocity
θ = 35°, the launch angle
Let m = the mass of the cannonball.
Let a = the acceleration of the ball when fired.
The net force acting on the ball is
F - mg sinθ = 2000 - 9.8*m*sin(35°) = 2000 - 5.621*m N
Then, from Newton's Law of motion,
F = ma
2000 - 5.621*m = m*a (1)
The launch velocity is V₀ = 8.3 m/s, therefore
V₀² = 2*a*L
(83 m/s)² = 2*(a m/s²)*(2.41 m)
6889 = 4.82*a
a = 6889/4.82 = 1429.25 m/s² (2)
Insert (2) into (1)
2000 - 5.621*m = 1429.25*m
2000 = 1434.871*m
m = 1.394 kg
Answer: 1.394 kg