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inessss [21]
4 years ago
5

Please Help! ASAP!

Physics
1 answer:
posledela4 years ago
8 0

Answer:

C and D

Explanation:

Its possible to use telescopes during the day time.

Gamma rays were found on earth's atmosphere.

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Describe what happens to the energy of a wave if the frequency decreases and the frequency increased
Nata [24]

"As frequency increases, wavelength decreases. Frequency and wavelength are inversely proportional. This basically means that when the wavelength is increased, the frequency decreases and vice versa. Wavelength is described as the distance between a trough to a trough or a crest to a crest."

I'd recommend paraphrasing it tho.

4 0
3 years ago
Read 2 more answers
A force is required for any change in an object's motion to occur.<br><br> True<br> False
sertanlavr [38]

Answer:

True

Explanation:

an object in motion stays in motion unless acted upon by another force

7 0
3 years ago
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How much money did congress authorize to be spent on its construction
ser-zykov [4K]

Answer:

20,000

Explanation:

5 0
4 years ago
How many miles will be completed by a runner with a constant speed of 5 mph in 30 minutes?
Mrac [35]
Distance=speed times time
speed=5mph=5 \frac{mi}{hr}
time=30mins=30/60hr=1/2hr

so
distance=speed times time
distance=(5\frac{mi}{hr})(\frac{1}{2}hr)
distance=\frac{5mi \space\ hr}{2hr}
distance=2.5mi

he will run 2.5 miles in 30 mins at the speed of 5mph
5 0
3 years ago
New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in
seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

  • Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
  • As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

        v = \lambda * f  (1)

  • Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

       \lambda_{low} = \frac{c}{f_{high}}  = \frac{3e8m/s}{300e9Hz} = 1 mm (2)

      \lambda_{high} = \frac{c}{f_{low}}  = \frac{3e8m/s}{30e9Hz} = 10 mm (3)

4 0
3 years ago
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