Answer:
The magnitude of angular acceleration is 
.
Explanation:
Given that,
Initial angular velocity, 
When it switched off, it comes o rest, 
Number of revolution, 
We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :
So, the magnitude of angular acceleration is . Hence, this is the required solution.
Answer:
The canon B hits the ground fast.
Explanation:
Given that,
Speed of cannon A = 85 m/s
Speed of cannon B= 100 m/s
Speed of cannon C = 75 m/s
We need to calculate the cannonballs will hit the ground with the greatest speed
Using conservation of energy
The final kinetic energy of canon depends on initial kinetic energy and potential energy.
The final velocity depends upon initial velocity and initial height.
So, the initial velocity of canon B is high.
Hence, The canon B hits the ground fast.
Answer:
Net force: 20 N to the right
mass of the bag: 20.489 kg
acceleration: 0.976 m/s^2
Explanation:
Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:
195 N - 175 N = 20 N
So net force on the bag is 20 N to the right.
The mass of the bag can be found using the value of the weight force: 201 N:
mass = Weight/g = 201 / 9.81 = 20.489 kg
and the acceleration of the bag can be found as the net force divided by the mass we just found:
acceleration = 20 N / 20.489 kg = 0.976 m/s^2
It is converted to kinetic energy.
F = m₁ a₁ = m₂ a₂
if m₁ = m and m₂ = 2m :
F = ma₁ = 2m a₂ ⇒ a₁ = 2 a₂
since v = at + v₀ with t = 3, v₀ = 0 ⇒ v = 3a:
v₁ = 2 v₂
since p = vm with v₁ = 2v and v₂ = v :
p₁ = v₁m₁ = 2v ⁻ m
p₂ = v₂m₂ = v ⁻ 2m
p₁ = p₂
