Give an example of a change that the ecosystem was not able to recover from. Can you explain why? (gizmo)

Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths f

Bogdan [553]

Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t $_{min$ = ( m + 1/2) × λ/2n

we substitute

t $_{min$ = ( 0 + 1/2) × 711 /2(1.21)

t $_{min$ = 0.5 × 711/2.42

t $_{min$ = 0.5 × 293.80165

t $_{min$ = 146.9 nm

Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm

8 0

3 years ago

The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the poo

DedPeter [7]

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

$n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m$

or

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

4 0

3 years ago

Which of the following is an example of kinetic energy?

tresset_1 [31]

B. Sound, because everything else sits still and sound waves move

5 0

3 years ago

A research team developed a robot named Ellie. Ellie ran 1,000 meters for 200 seconds from the research building, rested for 100

Verizon [17]

Answer:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)

Explanation:

Attached you will find the plot of position vs time of Ellie´s movement.

The velocity is the displacement of the object over time relative to the system of reference. The speed, in change, is the traveled distance over time in disregard of the system of reference.

So, the velocity is calculated as follows:

v = Δx / Δt

where

Δx = final position - initial position

Δt = elapsed time

1) The average velocity of Ellie while running is:

v = 1000 m - 0 m / 200 s = 5 m/s

While resting:

v = 0 m - 0 m / 100 s = 0 m/s

And while walking back:

v = 0 m - 1000 m / 1000 s = - 1 m/s

Note that in this last case, the initial position is 1000 m because Ellie is 1000 m from the origin of the system of reference when she walks back. The final position will be the origin of the system of reference, 0 m.

Comparing with the graphic, the velocity is the slope of the function position(t).

Then:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

2) The speed is the distance traveled over time:

Running speed = 1000 m / 200 s = 5m /s

Resting speed = 0 m / 100 s = 0 m/s

Walking speed = 1000 m/ 1000 s = 1 m/s

Then:

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)

4 0

3 years ago

A person travels by car from one city to another with different constant speeds between pair of cities. She drives for 36 min at

Softa [21]

Change minutes to hrs, divide by 60:
30 min = .50 hrs
45 min = .75 hrs
12 min = .20 hrs
----------------
total + 1.45 hrs, total travel time
:

let a = average speed for the trip
:
Write a dist equation, dist = speed * time
:
80(.5) + 100(.20) + 40(.75) = 1.45a
40 + 20 + 30 = 1.45a
90 = 1.45a
a =
a = 62.069 km/h, for the entire trip
and
90 km is the total distance

3 0

3 years ago