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3 years ago

The amount of randomness of the atoms in a system is an indication of the

Physics

1 answer:

GuDViN [60]3 years ago

4 0

Answer:

Entropy of the system

Explanation: entropy measures the randomness

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Which biome has the most variable year round temperature

skad [1K]

That should be the Grasslands ?

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3 years ago

6. Explain the difference between a neutral atom and an ion?

Deffense [45]

Answer:

a neutral atom doesn't have a charge and an ion does.

Explanation:

When atoms bond they can gain or lose electrons which gives them a charge.

anion = negatively charged ion

cation = positively charged ion

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3 years ago

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Answer:

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3 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

5 0

3 years ago

A phonograph record 0.15 m in its radius rotates 18 times per 90 seconds what is the frequency?

ioda

Answer:

The frequency of the phonograph record is 0.2 Hz

Explanation:

The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period

The given parameters of the phonograph record are;

The radius of the record = 0.15 m

The number of times the phonograph record rotates, n = 18 times

The time it takes the phonograph record to rotate the 18 times, t = 90 seconds

The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)

∴ The frequency of the phonograph record, f = n/t = 18/(90 s) = 0.2 Hz

The frequency of the phonograph record = 0.2 Hz.

6 0

3 years ago