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3 years ago

Where are you most likely to build up enough static charge to receive a

2 answers:

6 0

You're most likely to build up enough static charge to receive a shock by walking around in a carpeted restaurant in the desert. (A)

Walking on carpet is the fastest way to accumulate charge, and the dry desert air prevents the charge from dribbling off of you and away.

When I walked on stones in the Sinai Desert, the dry wind with a little bit of sand or dust in it built up enough static charge on me that I got a shock every time I stood less than a foot away from my partner.

I had the same experience a few years later near Ouarzazate in the interior of Morocco.

When you hear people say "the desert is dry", they mean it's DRY !

gulaghasi [49]3 years ago

4 0

Answer:

A. In a carpeted resteraunt in the desert

Explanation:

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weathering occurs when rocks sit in a pool of saltw ater.

Explanation:

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2 years ago

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

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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 $\vartriangle r=m \lambda$