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bixtya [17]
3 years ago
13

ou are out stargazing with your 13.4-cm telescope. You point your telescope at an interesting formation in the sky, which you th

ink is a binary star system. A binary star system consists of two stars in orbit around each other. You guess that the average wavelength coming from the stars is 518 nm. What is the minimum angular separation between the two stars required for your telescope to resolve the two stars of the binary system
Physics
1 answer:
Alinara [238K]3 years ago
8 0

Answer:

θ = 4.716 10⁻⁶ rad

Explanation:

In order for the releases to be considered separate, they must meet the Rayleigh criterion that establishes that the maximum diffraction of one star must coincide with the first minimum of the diffraction pattern of the second star.

We use the diffraction equation for a slit

            a sin θ = m λ

The minimum occurs at m = 1

             sin θ = λ / a

Since the angles in these systems are very small, we can approximate the sine to its angle in radians

             θ = λ / a

The telescope has a circular aperture whereby polar cords should be used, which introduces a constant number

           θ = 1.22 λ / a

Let's calculate

          θ = 1.22 518 10⁻⁹ / 13.4 10⁻²

          θ = 4.716 10⁻⁶ rad

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Saturn isn’t the only planet with rings, but they are the most beautiful ones. <br> Fact or Opinion?
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In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike
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a) t=\sqrt{\frac{2h}{g}}

b) v=\frac{d}{\sqrt{\frac{2h}{g}}}

c) v=\sqrt{d^2(\frac{g}{2h})+(2gh)}

d) \theta=tan^{-1}(\frac{2h}{d}) (radians)

Explanation:

a)

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

s=ut+\frac{1}{2}at^2

where here:

s=h (the vertical displacement is the height of the counter)

u=0 (the initial vertical velocity of the cup is zero)

a=g (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}

b)

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

v=\frac{d}{t}

where

d is the distance covered

t is the time taken

Here, the distance covered is d, the distance from the base of the counter, while the time taken is the time of flight:

t=\sqrt{\frac{2h}{g}}

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

v=\frac{d}{\sqrt{\frac{2h}{g}}}

c)

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}

The vertical speed instead is given by the suvat equation:

v_y=u_y + at

where:

u_y=0 is the initial vertical velocity

a=g is the acceleration

t=\sqrt{\frac{2h}{g}} is the time of flight

Substituting,

v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}

d)

Just before hitting the floor, the velocity of the cup has two components:

v_x=d\sqrt{\frac{g}{2h}} is the horizontal component (in the forward direction)

v_y=\sqrt{2gh} is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

tan \theta = \frac{v_y}{v_x}

where here \theta is measured as below the horizontal direction.

Substituting the expressions for v_x,v_y, we find:

tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}

So

\theta=tan^{-1}(\frac{2h}{d}) (radians)

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Answer:

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Now in present experiment Rutherford found that very few alpha particles are bounced back along same path which shows that very small region inside the nucleus is having positive charge and rest part of the atom is empty.

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so correct answer will be

1. The nucleus is about 1/2 the size of the atom

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3 years ago
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