The equation shows neutralization of an acid and a base to produce a salt and water.
2 answers:
You might be interested in
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