a. What is the absorption coefficient for tin, with an imaginary part of the refractive index equal to 5.3 for 589.3-nm light? b
. At what depth is 99% of normally incident sodium light absorbed in tin? How thin would the metal Tin have to be for 50% of the light to come out the other side?
1 answer:
You might be interested in
Newton's second law allows us to find the force of the block on the table is 126 N
Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body
Σ F = m a
Where the bold letters indicate vectors, m is the mass and the acceleration of the body
A free body diagram is a diagram where the forces are represented without the details of the bodies, in the attached we can see a free body diagram of the system.
Let's start by finding the acceleration of the elevator with kinematics
v = v₀ + a t
a =
Where v and v₀ are the current and initial velocity, respectively, at acceleration and t is the time
a =
a = 3.5 m / s²
Let's write Newton's second law for each body
The book
N₂ - W₂ + N₁ = m a
Table
N₁ - W₁ - W₂ = M a
W₁ = Mg
W₂ = mg
N₁ = (M + m) g + M a
N₁ = (10 + 2) 9.8 + 10 3.5
N₁ = 152.6 N
This is the reaction of the earth to the support of the block and the table
N₂ = ma + m g - N₁
N₂ = m ( a +g) - N₁
N₂ = 2 (3.5 + 9.8) - 152.6
N₂ = 26.6 - 152.6
N₂ = -126 N
The negative sign indicates that the direction is opposite to the one assigned, this is the action of the block on the table.
In conclusion using Newton's second law we can find the forces of the block on the table is 126 N
Learn more here: brainly.com/question/19860811
Answer:
given,
speed of the car = 20 m/s
final speed of car = 0 m/s
distance between car and the deer = 38 m
reaction time, t = 0.5 s
deceleration of the car = 10 m/s².
a) distance between deer and car
distance travel in the reaction time
d₁ = v x t
d₁ = 20 x 0.5 = 10 m
distance travel after you apply brake
using equation of motion
v² = u² + 2 a s
0 = 20² - 2 x 10 x s
s = 20 m
total distance traveled by the car
D = d₁ + d₂
D = 20 + 10 = 30 m
distance between car and the deer = 38 m - 30 m
= 8 m
b) now, maximum speed car.
distance travel in reaction time
d₁ = s x t
d₁ = 0.5 V
distance left between them
d₂ = 38 - d₁
d₂ = 38 - 0.5 V
distance travel after you apply brake
using equation of motion
v² = u² + 2 a d₂
0 = (V)² - 2 x 10 x (38 - 0.5 V)
V² + 10 V - 760 = 0
now, solving the quadratic equation
V = 23.01 , -33.01
rejecting the negative term.
hence, maximum speed of the car could be V = 23.01 m/s