Answer:
- the coating’s index of refraction is 1.25
- the required thickness is 104.1667 nm
Explanation:
Given the data in the question;
Thickness of coating t = 100 nm
wavelength λ = 500nm
we know that refractive index is;
t = λ/4n
make n, the subject of formula
t4n = λ
n = λ / 4t
we substitute
n = 500 / ( 4 × 100 )
n = 500 / 400
n = 1.25
Therefore, the coating’s index of refraction is 1.25
2)
given that;
Index of refraction of the coating; n = 1.20
λ = 500 nm
thickness of coating t = ?
t = λ / 4n
we substitute
t = 500 / ( 4 × 1.2 )
t = 500 / 4.8
t = 104.1667 nm
Therefore, the required thickness is 104.1667 nm
Answer:
Speed of air = 1106.38 ft/s
Explanation:
Speed of sound in air with temperature
Here speed is in m/s and T is in celcius scale.
T = 50°F
Substituting
Now we need to convert m/s in to ft/s.
1 m = 3.28 ft
Substituting
Speed of air = 1106.38 ft/s
Answer: he did travel 15 meters.
Explanation:
We have the data:
Acceleration = a = 1.2 m/s^2
Time lapes = 3 seconds
Initial speed = 3.2 m/s.
Then we start writing the acceleration:
a(t) = 1.2 m/s^2
now for the velocity, we integrate over time:
v(t) = (1.2 m/s^2)*t + v0
with v0 = 3.2 m/s
v(t) = (1.2 m/s^2)*t + 3.2 m/s
For the position, we integrate again.
p(t) = (1/2)*(1.2 m/s^2)*t^2 + 3.2m/s*t + p0
Because we want to know the displacementin those 3 seconds ( p(3s) - p(0s)) we can use p0 = 0m
Then the displacement at t = 3s will be equal to p(3s).
p(3s) = (1/2)*(1.2 m/s^2)*(3s)^2 + 3.2m/s*3s = 15m
The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.
Answer:
D. At x=0, it's acceleration is at a maximum
Explanation:
As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.
Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.
From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.
The same can be said as the box travels backward from point A to -A
a substance's density is the same at a certain pressure and temperature, and the density of one substance is usually different than another substance.
