Answer: from the information given, the velocity of the water will decrease but the pipe size will remain the same.
This can be proved with bernoulli's equation.
Explanation: careful analysis of the system using bernoulli's equation of flow is shown in the image attached
Answer:
The total energy of the composite system is 7.8 J.
Explanation:
Given that,
Height = 0.15 m
Radius of circular arc = 0.27 m
Suppose, the entire track is friction less. a bullet with a m₁ = 30 g mass is fired horizontally into a block of wood with m₂ = 5.29 kg mass. the acceleration of gravity is 9.8 m/s.
Calculate the total energy of the composite system at any time after the collision.
We need to calculate the total energy of the composite system
Total energy of the system at any time = Potential energy of the system at the stopping point
Put the value in to the formula
Hence, The total energy of the composite system is 7.8 J.
Answer:
0.027648 kgm²
Explanation:
M = Mass of disc = 1.2 kg
r = Radius of disc = 0.16 m
m = Mass of rod = 0.16 kg
R = Rod distance = 0.16 m
Moment of inertia of disk is given by
Moment of inertia of the three rods
The total moment of inertia is given by
The moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center is 0.027648 kgm²
Answer:
71.76 m
Explanation:
We will solve this question using the work energy theorem.
The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.
ΔK.E = W
In the attached free body diagram for the question, the forces acting on the puck are given.
ΔK.E = (final kinetic energy) - (initial kinetic energy)
Final kinetic energy = 0 J (since the puck comes to a stop)
Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J
ΔK.E = 0 - 67.6 = - 67.6 J
W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)
Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h
Workdone by the frictional force = F × d
F = μ N
μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)
N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N
F = μ N = 0.3 × 1.697 = 0.509 N
where d = distance along the incline that the puck travels.
d = h/sin 30° = 2h (from trigonometric relations)
Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h
ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)
- 67.6 = - 1.96h + 1.02h
-0.942h = - 67.6
h = 71.76 m
