Answer:
Explanation:
As we know that the combination is maintained at rest position
So we will take net torque on the system to be ZERO
so we know that
here we will have
so we have
so we have
The dens or the odontoid process of the axis or the second cervical spine forms a pivot point with the atlas or the first cervical vertebrae that is responsible for the nodding and the rotational movements of the head. This is reinforced by ligaments and the atlanto-occipital joint that allows the head to make a nodding or up and down movement on the vertebral column.
Answer:
Explanation:
a )
The stored elastic energy of compressed spring
= 1 / 2 k X²
= .5 x 19.6 x (.20)²
= .392 J
b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .
c ) Let h be the distance along the incline which the block ascends.
vertical height attained ( H ) =h sin30
= .5 h
elastic potential energy = gravitational energy
.392 = mg H
.392 = 2 x 9.8 x .5 h
h = .04 m
4 cm .
=
Answer:
a)- 1.799 rad/sec²
b)- 17.6 x 10ˉ³Nm
Explanation:
ω₀ = 720 rev/min x (1 min/60 sec) x (2π rad / 1 rev) = 24π rad/s
a) Assuming a constant angular acceleration, the formula will be
α = (ωf -ω₀) / t
As final state of the grindstone is at rest, so ωf =0
⇒ α = (0-24π) / 41.9 = - 1.799 rad/sec²
b)Moment of inertia I for a disk about its central axis
I = ½mr²
where m=2kg and radius 'r'= 0.099m
I = ½(2)(0.099²)
I = 9.8 x 10ˉ³ kgm²
Next is to determine the frictional torque exerted on the grindstone, that caused it to stop, applying the rotational equivalent of the Newton's 2nd law:
τ = I α =>(9.8 x 10ˉ³)(- 1.799)
τ = - 17.6 x 10ˉ³Nm
(The negative sign indicates that the frictional torque opposes to the rotation of the grindstone).
kinetic energy is given as
KE = (0.5) m v²
given that : v = speed of the bottle in each case = 4 m/s
when m = 0.125 kg
KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J
when m = 0.250 kg
KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J
when m = 0.375 kg
KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J
when m = 0.0.500 kg
KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J
