When the medium is uniform, how do light waves pass through it?
1 answer:
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Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 × m
dynamic viscosity = 1.75 × Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ
so
= µ ............1
put here value
= 1.75× ×
= 0.175 v
and
area between air and puck is given by
Area =
area =
area = 7.85 × m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 ×
force = 1.374 × v
and now apply newton second law
force = mass × acceleration
- force =
- 1.374 × v =
t =
time = 2.18
so time required after impact for a puck is 2.18 seconds
Answer:
Given that
m = 5.3 kg
Fx = 2x + 4
We know that work done by force F given as
w= ∫ F. dx
a)
Given that x=1.08 m to x=6.5 m
Fx = 2x + 4
w= ∫ F. dx
w=62.7 J
b)
We know that potential energy given as
∫ dU = -∫F.dx ( w= ∫ F. dx)
ΔU= -62.7 J
c)
We know that form work power energy theorem
Net work = Change in kinetic energy
W= KE₂ - KE₁
62.7 =KE₂ - (1/2)x 5.3 x 3²
KE₂ = 86.55 J
This is the kinetic energy at 6.5m