Answer:
Explanation:
thickness of oil t = 200 nm
index of refraction μ = 1.5
For transmitted light :---
path difference = 2μ t
For constructive interference
path difference = n λ , λ is wavelength of light
2μ t = n λ
λ = 2μ t / n
For longest λ , n = 1
λ = 2μ t
= 2 x 1.5 x 200 nm
= 600 nm
Wavelength in water
= 600 / refractive index of water
= 600 / 1.33
= 451.1 nm Ans
The woman on the platform is correct because it is the pace of the man moving on the train not walking.
Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight
Answer:
S=48.29 m
Explanation:
Given that the height of the hill h = 2.9 m
Coefficient of kinetic friction between his sled and the snow μ = 0.08
Let u be the speed of the skier at the bottom of the hill.
By applying conservation of energy at the top and bottom of the inclined plane we get.
Potential Energy=kinetic Energy
mgh = (1/2) mu²
u² = 2gh
u²=2(9.81)(2.9)
=56.89
u=7.54 m/s
a = - f / m
a = - μ*m*g / m
a = - μg
From equation of motion
v²- u² = 2 -μ g S
v=0 m/s
-(7.54)²=-2(0.06)(9.81)S
S=48.29 m
