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3 years ago

An area where the particles in a medium are spaced close together is called a _____.

Physics

2 answers:

Nina [5.8K]3 years ago

6 0

the answer is B compression

Alex17521 [72]3 years ago

3 0

Answer:

compression

Explanation:

Gradpoint

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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

Compare two processes we use to measure the age of Earth and its parts.

yan [13]

Answer:

Generally speaking, scientists have developed four different methods of determining the age of the earth. By using these methods, or a combination of them, the age of geological formations created by past events and even the fossilized bones of prehistoric animals can be determined.

Based on the research done by scientists, the Geologic Column, a graph that illustrates earth age, was developed. The Column, with its names for the epochs and eras of time, illustrates what scientists think was taking place in earth history. A small version of the Geologic Column is pictured at the right.

Explanation:

I hope this helps :)

3 0

3 years ago

If a certain silver wire has a resistance of 3.00 Ï at 11.0Â°c, what resistance will it have at 25.0Â°c?

Romashka [77]

11.30 times the nuber of toys

7 0

3 years ago

It takes 6400 years for one gram of radium to decay away to only 1/16 (one-sixteenth) of a gram. The half-life of radium is

Vsevolod [243]

1/16........................................

4 0

3 years ago

2.5 g of helium at an initial temperature of 300 K interacts thermally with 9.0 g of oxygen at an initial temperature of 620 K .

muminat

Answer:

Explanation:

2.5 g of He = 2.5 / 4 mole

= .625 moles

9 g of oxygen = 9/32

= .28 mole of oxygen

C_p of He = 3/2 R

C_p of O₂ = 5/2 R

A ) Initial thermal energy of He = 3/2 n R T

= 1.5 x .625 x 8.32 x 300

= 2340 J

Initial thermal energy of O₂ = 5/2 n R T

= 2.5 x .28 x 8.32 x 620

= 3610.88 J

B ) If T be the equilibrium temperature after mixing

gain of heat by helium

= n C_p Δ T

= .625 x 3/2 R x ( T - 300 )

Loss of heat by oxygen

n C_p Δ T

= .28 x 5/2 R x ( 620 - T )

Loss of heat = gain of heat

.625 x 3/2 R x ( T - 300 ) = .28 x 5/2 R x ( 620 - T )

1.875 T- 562.5 = 868- 1.4 T

3.275 T = 1430,5

T = 436.8 K

Thermal energy of He

= 1.5 x .625 x 8.32 x 436.8

= 3407 J

thermal energy of O₂

= 2.5 x .28 x 8.32 x 436.8

= 2543.92 J

C )

Heat energy transferred

= .28 x 5/2 R x ( 620 - T )

= .28 x 5/2 x 8.32 x ( 620 - 436.8 )

1066.95 J

Heat will flow from O₂ to He

Final temperature is 436.8 K

7 0

3 years ago