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Physics

1 answer:

Ratling [72]3 years ago

8 0

Answer:

1. 1816

2. In order to make a weather map, what process was followed? They had to manually plot out the points of equal atmospheric pressure on an Isobar. Then, plot the rest of the information intended for that map.

3. Early weather maps could only be surface charts. There was no way of knowing what was happening higher in the troposphere so a complete picture of the atmosphere was impossible.

4. The weather maps are the most valuable tool to a meteorologist because it helps with the weather forecast. Without the weather maps it would be hard to predict what the weather would be. They help summarize what is going on in the atmosphere.

5. The information is then electronically sent to Washington, DC where it is analyzed.

Hope it helps! <3

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It is 32 degrees F outside. What is this in Kelvin?

igomit [66]

Answer:

273.15

Explanation:(32°F − 32) × 5/9 + 273.15 = 273.15K

3 0

3 years ago

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen

Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}$

$sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m$