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SpyIntel [72]
3 years ago
6

Post-Impressionist Georges Seurat is famous for a painting technique known as pointillism. His paintings are made up of differen

t-colored dots, and each dot is roughly 2 mm in diameter. When viewed from far enough away, the dots blend together and only large-scale images are seen. Assuming that the human pupil is about 2.9 mm in diameter, determine an approximate minimum distance you need to stand from one of Seurat's paintings to see the dots blended together. (Assume two dots touch, such that their center-to-center distance is equal to one diameter. Also assume light with a wavelength of 563 nm.)
Physics
1 answer:
algol [13]3 years ago
6 0

Answer:

84 cm

Explanation:

The resolving power of the eye = 1.22 λ / D where λ is wavelength of light and D is diameter of the eye .

Putting the values in the formula

The resolving power of the eye = 1.22 λ / 2.9 x 10⁻³

Let the minimum distance you need to stand from one of Seurat's paintings to see the dots blended together  be D .

d = distance between two dots = diameter of dot = 2 x 10⁻³ m .

Given ,

λ D / d =  1.22 λ / 2.9 x 10⁻³

D /2 x 10⁻³   =  1.22  / 2.9 x 10⁻³

D = 2 x 1.22 / 2.9 m

= .84 m

= 84 cm .

minimum distance you need to stand from one of Seurat's paintings to see the dots blended together is 84 cm .

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Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea
Setler [38]

Answer:

a)  M = 4,997 10²⁰ kg ,  b)   T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

          v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

         a = (v₀ - v) / t

         a = (15 - (-15)) /9.00 = 30/9

         a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

          F = m a

         G m M / r² = m a

         M = a r² / G

Let's calculate

         M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

         M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

         R = R_{m} + h

         R = 1 10⁵ + 2.00 10⁴

         R = 12 10⁴ m

         F = m a

        G m M / R² = m a

Centripetal acceleration is

         a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

        d = 2π R

        v = d / t

        v = 2π R / T

Let's replace

        G m M / R² = m (2π R / T)² / R

        G M = R³ 4π² / T²

        T² = 4π² R³ / G M

       T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

       T² = 6.82 10¹⁶ / 3.33 10¹⁰

       T = √ (2,048 10⁶)

       T = 1.43 10³ s

3 0
3 years ago
A car is traveling at a velocity of 22 m/s when the driver puts on the brakes
Brums [2.3K]

The car’s velocity at the end of this distance is <em>18.17 m/s.</em>

Given the following data:

  • Initial velocity, U = 22 m/s
  • Deceleration, d = 1.4 m/s^2
  • Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

V^2 = U^2 + 2dS

Substituting the values into the formula, we have;

V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}

<em>Final velocity, V = 18.17 m/s</em>

Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>

<em></em>

Read more: brainly.com/question/8898885

8 0
2 years ago
How long would it take for Sofia to walk 300 meters if she is walking at a velocity of 2.5 m/s?
Vaselesa [24]

Answer:

Time=120seconds

Explanation:

S=300m

V=2.5m/s

t=?

V=S/t

t=S/V

t=300/2.5

t=120 second

7 0
3 years ago
Read 2 more answers
Which of the following is NOT common of elite Shang burials?
Zigmanuir [339]

Answer:

Large above ground mausoleums were not common in the elite Shang burials.

Explanation:

Large, above the ground mausoleums were not common so the answer is option B.

6 0
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Rudik [331]

Explanation:

1. To graphically add vectors, use the tail-to-tip method.  Draw the first vector (it doesn't matter which), then draw the second vector where the first vector ends.  The resultant vector is from the tail of the first vector to the tip of the second vector.

This graph shows two ways to get the resultant: A + B or B + A.

desmos.com/calculator/bqhcclhhqc

2. To algebraically add vectors, split each vector into x and y components.

Aₓ = 5.0 cos 45 = 3.5

Aᵧ = 5.0 sin 45 = 3.5

Bₓ = 2.0 cos 180 = -2.0

Bᵧ = 5.0 sin 180 = 0

The components of the resultant vector are the sums of the components of A and B.

Cₓ = 3.5 + -2.0 = 1.5

Cᵧ = 3.5 + 0 = 3.5

The magnitude of the resultant vector is found with Pythagorean theorem, and the direction is found with tangent.

C = √(Cₓ² + Cᵧ²) ≈ 3.9 m/s

θ = atan(Cᵧ / Cₓ) ≈ 67°

4 0
3 years ago
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