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larisa86 [58]
3 years ago
7

HELP NEEDED

Physics
1 answer:
MrMuchimi3 years ago
8 0

Answer:

D the development of an ozone layer

Explanation:

the rest don't relate to the atmosphere

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"What kind of electromagnetic radiation can be detected by a human body? I) visible light II) infrared light III) ultra-violet l
Vinvika [58]

Answer:

7. I and III only.

Explanation:

Only portions of the electromagnetic spectrum is visible to the human eye. Electromagnetic radiation in this range of wavelengths is called visible light which have wavelengths from about 390 to 750 nm that is; ultraviolet light and visible light.

4 0
3 years ago
A hydrogen atom in the n=7 state decays to the n=4 state. what is the wavelength of the photon that the hydrogen atom emits? use
frez [133]

A hydrogen atom in the n=7 state decays to the n=4 state. The wavelength of the photon that the hydrogen atom emits is 4592.59nm.

The Energy of photon is the energy possessed by a photon when it moves from a high energy level to a low energy level. It emits a photon of a certain wavelength. The following relation can be used to find out the relation between the energy levels and the energy possessed:

E = 13.6 × Z² (1/n₂² - 1/n₁²) eV

where, n₁ is the initial energy level i.e. n₁ =7

            n₂ is the higher energy level i.e. n₂ = 4

            E is the energy possessed

            Z is the atomic number, Z = 1 for H-atom

Subsituting in above equation,

E = 13.6 (1/16 - 1/49) eV

E = 0.27 eV

We know that,

E = hc / λ  

where, h is Planck constant

           c is speed of light

            λ is wavelength

On subsituting,

0.27 eV = 1240/ λ

⇒ λ = 4592.59 nm

Hence, the wavelength of photon emitted by Hydrogen atom is 4592.59nm.

Learn more about Energy of Photon here, brainly.com/question/2393994

#SPJ4

 

5 0
2 years ago
Help, please<br> and explain
11Alexandr11 [23.1K]
We could tell a force is acting on an object if the object is being pushed, pulled, or moved in any way
5 0
4 years ago
Which type of mechanical wave is a sound wave?
kykrilka [37]
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4 0
3 years ago
Read 2 more answers
Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19
Alisiya [41]

Answer:

The workdone is  W = 9.28 * 10^{3} J

Explanation:

From the question we are told that

   The height of the cylinder is  h = 0.588\ m

   The face Area is  A = 4.19 \ m^2

    The density of the cylinder is \rho  =  0.346 * \rho_w

     Where \rho_w is the density of freshwater which has a constant value

              \rho_w = 1000 kg/m^3

     

Now  

     Let the final height of the device under the water be  =  h_f

      Let  the initial volume underwater be = V_n

     Let the initial height under water be  = h_i

      Let the final volume under water be  = V_f

According to the rule of floatation

        The weight of the cylinder =  Upward thrust

This is mathematically represented as

          \rho_c g V_n = \rho_w gV_f

         \rho_c A h = \rho A h_f

So      \frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}

   =>     \frac{h_f}{h_c}  = 0.346

Now the work done is mathematically represented as  

          W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh

               =   \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.

              = \frac{g A \rho}{2}  [h^2 - h_f^2]

              = \frac{g A \rho}{2} (h^2)  [1  - \frac{h_f^2}{h^2} ]

Substituting values

        W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)

        W = 9.28 * 10^{3} J

4 0
3 years ago
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