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2 years ago

Copernicus and other astronomers before him thought that celestial bodies followed a _____ orbital path.

2 answers:

6 0

The correct answer is circular. Copernicus and other astronomers before him thought that celestial bodies followed a circular orbital path. Copernicus was a Polish astronomer that concluded that the sun is at rest near the center of the universe and the earth is revolving around it annually. This theory is called heliocentric.

Alexus [3.1K]2 years ago

5 0

the answer is circular

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An FM radio station broadcasts electromagnetic radiation at a frequency of 93.5 MHz. The wavelength of this radiation is _______

Alborosie

Answer:

3.21

Explanation:

The relation between frequency and wavelength is shown below as:

$c=frequency\times Wavelength
$

c is the speed of light having value $3\times 10^8\ m/s$

Given, Frequency = 93.5 MHz = $93.5\times 10^{6}\ Hz$

Thus, Wavelength is:

$Wavelength=\frac{c}{Frequency}$

$Wavelength=\frac{3\times 10^8}{93.5\times 10^{6}}\ m$

$Wavelength=3.21 \ m$

Answer - A.

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3 years ago

Rita conducts an experiment on how the amount of precipitation each fall affects

hodyreva [135]

Answer:

Explanation:

As spring season is a yearly phenomenon so, Rita should organize her data on yearly basis. Firstly, she should plan the procedure of her experiment and collect the data according to it. Secondly, identify the attribute of each object of her experiment. Thirdly, she can organize and segregate her data in tabular form, graphical form or diagrammatically.

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3 years ago

The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation

makvit [3.9K]

Answer:

Part A:

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

Part B:

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

Part C:

a) If the distance to the screen is decreased the fringe spacing will decrease.

Part D:

The dot in the center of fringe E is $920\ x\ 10^{-9} m$ farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

$\vartriangle y =\frac{m \lambda L}{d}$

$m=0,\pm 1,\pm 2,\pm 3,.....$

m is the order number.
$\lambda$ is the wavelength of the monochromatic light.
L is the distance between the screen and the two slits.
d is the distance between the slits.

Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light $\lambda$ is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.

Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.

Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be $\vartriangle r=m \lambda$