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Svet_ta [14]
4 years ago
7

dvanced photodiode detectors have a second light-emitting diode, operating at a wavelength of 2.0 × 10–7 m, to detect even small

er smoke particles from smoldering flames. What is the frequency difference between the two light beams?
Physics
1 answer:
ICE Princess25 [194]4 years ago
5 0

Answer:

1.5\cdot 10^{15}Hz

Explanation:

The relationship between wavelength and frequency is given by:

c=f \lambda

where

c=3\cdot 10^8 m/s is the speed of light

f is the frequency

\lambda is the wavelength

In this problem, we know the wavelength:

\lambda=2.0\cdot 10^{-7} m

So we can re-arrange the formula above to find the frequency:

f=\frac{c}{\lambda}=\frac{3.0\cdot 10^8 m/s}{2.0\cdot 10^{-7} m}=1.5\cdot 10^{15}Hz

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A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2000 kg SUV traveling from ea
skad [1K]

Answer:

v = 11.1m/s

Vx = 7.19m/s

Vy = 8.45m/s

V(sedan) = 30.4m/s

V(suv) = 25.9m/s

Explanation:

4 0
3 years ago
T=2pi square root 1/g solve for g.<br> Explanation would be really helpful.
Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}

Let me know if you have any questions.

7 0
3 years ago
What is the highest degrees above the horizon the moon ever gets during the year in the Yakima Valley ?
Ivahew [28]

The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.

Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima.  The latitude in the middle
of that intersection is 46.585° North.  <u>That's</u> the number we need.

Here's how I would do it:

-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.

-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°. 

-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.

This sets the limit of the highest in the sky that the moon can ever appear.

90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .

That doesn't happen regularly.  It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).

Depending on the time of year, that can be any time of the day or night.

The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.

In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky.  Then it's going to be somewhere near
67° above the horizon at midnight.


5 0
3 years ago
Clutter is _______. annoying frustrating death
Sindrei [870]

Answer: _____= beautiful, yet annoying frustrating death

Explanation:

3 0
3 years ago
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A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w
valkas [14]

Op here is another problem exactly like that. Just plug in your variables instead. And remember, time is never negative.

5 0
3 years ago
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