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o-na [289]
2 years ago
6

Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth

to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz. Calculate the wavelength of the infrared radiation.
Physics
1 answer:
Semmy [17]2 years ago
5 0

Answer:

\lambda=6.83\times 10^{-5}\ m

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m

So, the wavelength of the infrared radiation is 6.83\times 10^{-5}\ m.

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John accidentally drops his keys off the balcony at his apartment. John's friend Tony just happens to walk by at that moment and
EleoNora [17]

Explanation:

In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:

H = V²Sin²θ/2g

where,

V = Launch Speed = 18 m/s

θ = Launch Angle = 40°

g = 9.8 m/s²

Therefore,

H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)

H = 6.83 m

Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).

Therefore, the keys will make it back to John.

7 0
3 years ago
What is the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b?
xeze [42]

The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

<h3>Change in energy level of the electron</h3>

When photons jump from a higher energy level to a lower level, they emit or radiate energy.

The change in energy level of the electrons is calculated as follows;

ΔE = Eb - Ef

ΔE = -2.68 eV - (-5.74 eV)

ΔE = 3.06 eV

Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

Learn more about energy level here: brainly.com/question/14287666

#SPJ1

7 0
2 years ago
Click to review the online content. Then answer the question(s) below, using complete sentences. Scroll down to view additional
Nostrana [21]

Answer:

The general shape of a frequency distribution. For many data sets, statisticians use this information to determine whether there is a “normal” distribution of values. In normal distributions, the mean, median, and mode are the same. Whether the distribution is symmetrical or skewed in a certain direction. If the data is skewed to the right, this shows the mean will be greater than the median. Similarly, if the data is skewed left, the mean will be less than the median. The symmetry, or asymmetry, of the chart can help statisticians calculate probability. The modality of the data set. This means how many peaks exist in the data. For normal distributions, there will be one peak, or mode, in the data set.

Explanation:

i just got it right on edgenuity :)

6 0
3 years ago
Which occurs at a transform boundary?
andreev551 [17]
The answer is B. One plate slides past another. 

The San Andreas Fault in California and the Alpine Fault in New Zealand are examples of transform boundaries. 

Hope this helps! :)
5 0
3 years ago
What net force is required to accelerate a car at a rate of 2 m/s2 if the car has a mass of 3,000 kg?
Aleksandr [31]
The answer is 3000 × 2 = 6000 N
5 0
2 years ago
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