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Nat2105 [25]
3 years ago
14

Find the minimum thickness (in nm) of a soap bubble that appears green when illuminated by white light perpendicular to its surf

ace. Take the wavelength to be 549 nm, and assume the same index of refraction as water (nw

Physics
1 answer:
AnnyKZ [126]3 years ago
7 0

Answer:

103nm

Explanation:

Pls see attached file

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Find the time it takes for each object to accelerate from 0m/s to 40 m/s when pushed with 100N of force
Ivan

Answer:

40s

Explanation:

Given:

F=100N

V=40m/s and 0

force=change in momenton

F=mv-mu

t=40(100-0)/100

t=40/1

t=40s

4 0
3 years ago
The diagram below shows a food web.
Mekhanik [1.2K]
i think it’s B. sorry if i’m wrong
8 0
2 years ago
Read 2 more answers
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

5 0
3 years ago
g stAn experienced spear fisherman sees a small fish swimming in a tidal pool. If the fisherman sees the fish at approximately a
Luba_88 [7]

Answer:

  θ = 28.9

Explanation:

For this exercise let's use the law of refraction

          n₁ sin θ₁ = n₂ sin θ₂

where we use index 1 for air and index 2 for water where the fish is

        sin θ₂ = n₁ / n₂ sin θ₁

in this case the air repair index is 1 and the water 1.33

we substitute

        sin θ₂ = 1 / 1.33 sin t 40

        sin θ = 0.4833

         

        θ = sin⁻¹ 0.4833

        θ = 28.9

5 0
3 years ago
QUESTION 1
mario62 [17]
It's D. If it absorbed it would be turning to steam. I am taking honors chem in high school we are learning this.
7 0
3 years ago
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