Nuclide X has a higher rate of decay than nuclide Y. Based on this information, which of the following statements must be true?

It would be impossible to build a microscope that could use visible light to see the molecular structure of a crystal because. I

Serhud [2]

Answer:

the correct one is: a diffraction limits the resolving power to approximately the size of the wavelength of the light used

Explanation:

To be able to solve two structures with a light source, the Rayleigh criterion must be met that stable the two structures are solved when the first minimum of diffraction at one point is in the code of the first maximum of the other point

Using this criterion we can find an expression for the first minimization of the diffraction spectrum m = 1

sin θ tea = λ / a

now the structure of the comatose has a separation of around 1 nm and the wavelength of visible light ranges from 400 to 700 nm, when substituting we find

sin θ = 400/1 10

sin θ = 400

sin θ = 700/1

sin θ = 700

These values are neither impossible since the sin function is bounded between -1 to 1, so we cannot see the diffraction

When reviewing the different statements, the correct one is: a diffraction limits the resolving power to approximately the size of the wavelength of the light used:

6 0

2 years ago

The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p

Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

$4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$3.081x10^{16}m$

(c) How many meters in a light-year?

$9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$63325AU$

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

Since p is small it can be represent as:

$p(rad) = \frac{1AU}{d}$ (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

$p(rad) = \frac{p('')}{2.06x10^5}$ (2)

Then, equation 2 can be replace in equation 1:

$\frac{p('')}{2.06x10^5} = \frac{1AU}{d}$

$\frac{d}{1AU} = \frac{2.06x10^5}{p('')}$ (3)

From equation 3 it can be see that $1pc = 2.06x10^5 AU$

a) How many parsecs are there in one astronomical unit?

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

(c) How many meters in a light-year?

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

$x = (3x10^{8}m/s)(31536000s)$

$x = 9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

(e) How many light-years in a parsec?

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

5 0

3 years ago

Question 11 of 11 | Page 11 of 11

KiRa [710]

Answer:

Decreases the time period of revolution

Explanation:

The time period of Cygnus X-1 orbiting a massive star is 5.6 days.

The orbital velocity of a planet is given by the formula,

v = √[GM/(R + h)]

In the case of rotational motion, v = (R +h)ω

ω = √[GM/(R + h)] /(R +h)

Where 'ω' is the angular velocity of the planet

The time period of rotational motion is,

T = 2π/ω

By substitution,

T = 2π(R +h)√[(R + h)/GM]

Hence, from the above equation, if the mass of the star is greater, the gravitational force between them is greater. This would reduce the time period of revolution of the planet.

3 0

3 years ago

What is the purpose of a free body diagram

Amiraneli [1.4K]

A free body diagram is a representation of how the forces that are acting on a point or particle interact. You place your point at the origin and then draw your forces.

4 0

3 years ago

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The hydrocarbon C2H4 is a member of the_series.

Darina [25.2K]

Answer:

alkene series

Explanation:

the alkene series are the hydrocarbons e.gc2h4 c3h8

8 0

3 years ago

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