GIVING BRAINLIEST PLEASE HELP ME!!!
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Hi there!
We must begin by converting km/h to m/s using dimensional analysis:
Now, we can use the kinematic equation below to find the required acceleration:
vf² = vi² + 2ad
We can assume the object starts from rest, so:
vf² = 2ad
(17.22)²/(2 · 75) = a
a = 1.978 m/s²
Now, we can begin looking at forces.
For an object moving down a ramp experiencing friction and an applied force, we have the forces:
Fκ = μMgcosθ = Force due to kinetic friction
Mgsinθ = Force due to gravity
A = Applied Force
We can write out the summation. Let down the incline be positive.
ΣF = A + Mgsinθ - μMgcosθ
Or:
ma = A + Mgsinθ - μMgcosθ
We can plug in the given values:
22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))
A = 46.203 N
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