Answer:
Explanation:
a ) Conservation of momentum is followed
m₁ v₁ = m₂ v₂
3m x 2 = m v
v = 6 m/s
Total kinetic energy
= 1/2 x .35 x 6 ² + 1/2 x 1.05 x 2 ²
= 8.4 J
This energy must be stored as elastic energy in the spring which was released as kinetic energy on burning the cord.
Yes , the conservation of momentum will be followed in the bursting apart process. Only internal forces have been involved in the process. Two equal and opposite internal forces are created by spring which creates motion and generates kinetic energy.
It is False it's about how hard you work
Answer:
a) variation of the energy is equal to the work of the friction force
b) W = Em_{f} -Em₀
, c) he conservation of mechanical energy
Explanation:
a) In an analysis of this problem we can use the energy law, where at the moment the mechanical energy is started it is totally potential, and at the lowest point it is totally kinetic, we can suppose two possibilities, that the friction is zero and therefore by equalizing the energy we set the velocity at the lowest point.
Another case is if the friction is different from zero and in this case the variation of the energy is equal to the work of the friction force, in value it will be lower than in the calculations.
b) the calluses that he would use are to hinder the worker's friction force and energy
W = Em_{f} -Em₀
N d = ½ m v² - m g (y₂-y₁)
y₂-y₁ = 35 -10 = 25m
c) if there is no friction, the physical principle is the conservation of mechanical energy
If there is friction, the principle is that the non-conservative work is equal to the variation of the energy
Answer:
0.42 m/s²
Explanation:
r = radius of the flywheel = 0.300 m
w₀ = initial angular speed = 0 rad/s
w = final angular speed = ?
θ = angular displacement = 60 deg = 1.05 rad
α = angular acceleration = 0.6 rad/s²
Using the equation
w² = w₀² + 2 α θ
w² = 0² + 2 (0.6) (1.05)
w = 1.12 rad/s
Tangential acceleration is given as
= r α = (0.300) (0.6) = 0.18 m/s²
Radial acceleration is given as
= r w² = (0.300) (1.12)² = 0.38 m/s²
Magnitude of resultant acceleration is given as
= 0.42 m/s²
To solve this problem we will resort to the concept of angle of incidence and refraction.
Since it is a reflection on a mirror, the angle provided for refraction will be equal to that of the incidence, that is, 25 °
The angle of reflation is always perpendicular to the surface so it is necessary to find the angle with respect to it.
Therefore the angle of the reflected beam of light made with the surface normal is 65°
