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ASHA 777 [7]
3 years ago
11

A passenger jet flying at cruising altitude is often traveling in the upper troposphere at approximately 725 km per hour (or 450

mph). Why don't passenger jets burn up as meteors do?
Physics
1 answer:
kobusy [5.1K]3 years ago
7 0

Explanation:

Objects like meteors travel at high rates of the speed and thus, possess large amounts of the kinetic energy.  As the meteors enter Mesosphere, they enter more dense segment of earth's atmosphere with the greater numbers of the gas molecules.  Meteors which encounter the large numbers of the gas molecules at the high rates of the speed and generate large amounts of the friction in process. As a result, enormous heat is built up. This build-up of heat  ultimately causes the fast-traveling meteors to burn up in atmosphere.

<u>While passenger jets travel at very small speed as compared with the speed of the meteors and don not burn. </u>

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<span>It starts to cool down in Arizona in mid-October.  The weather remains good in October- April and the heat tends to break in mid of October. The time when it is coldest in Arizona in December or January. However, it's too hot in the month of July and Rain becomes common in Spring season.</span>
7 0
3 years ago
The first law of thermodynamics states that energy can neither be created nor destroyed. If this is true then why are we always
riadik2000 [5.3K]

Answer:

The second law of thermodynamics states in an isolated system, the entropy (the amount of thermal energy that cannot be converted into mechanical work, also known as the amount of disorder) always increases, therefore, an isolated system always require an external input (new sources) of energy for there to be orderliness or for the available energy of the system to remain constant or increase

Explanation:

5 0
3 years ago
Potential difference is measured in which units?<br> volts<br> amps<br> currents<br> watts
viktelen [127]
It is Volt
First option
4 0
3 years ago
Read 2 more answers
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: =E−Ryn2 In this equation Ry stands for
skelet666 [1.2K]

Answer:

  λ = 162 10⁻⁷ m

Explanation:

Bohr's model for the hydrogen atom gives energy by the equation

         E_{n} = - k²e² / 2m (1 / n²)

Where k is the Coulomb constant, e and m the charge and mass of the electron respectively and n is an integer

The Planck equation

           E = h f

The speed of light is

          c = λ f

          E = h c /λ

For a transition between two states we have

          E_{n} - E_{m} = - k²e² / 2m (1 / n_{f}² -1 / n_{i}²)

           h c / λ = -k² e² / 2m (1 / n_{f}² - 1/ n_{i}²)

           1 / λ = (- k² e² / 2m h c) (1 / n_{f}² - 1/n_{i}²)

The Rydberg constant with a value of 1,097 107 m-1 is the result of the constant in parentheses

Let's calculate the emission of the transition

            1 /λ = 1.097 10⁷ (1/10² - 1/8²)

            1 / λ = 1.097 10⁷ (0.01 - 0.015625)

            1 /λ = 0.006170625 10⁷

            λ = 162 10⁻⁷ m

3 0
4 years ago
R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d
padilas [110]

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

Learn more about Schwarzschild radius here:

brainly.com/question/12647190

#SPJ10

3 0
2 years ago
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