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Setler79 [48]
3 years ago
14

Emission of light from an atom occurs when an electron

Physics
2 answers:
il63 [147K]3 years ago
8 0
When, it goes for lower orbital
OverLord2011 [107]3 years ago
3 0
When an electron goes to lower orbital from higher orbital.
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B) The wavelength is longer and the pitch is lower

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What is the area of a rectangle having dimensions of 9.8m and 12.7m?
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Area = 124.46

Explanation:

I HOPE THIS WAS HELPFUL

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3 years ago
Please solve this question ​
klemol [59]

Answer:

Given: Initial Velocity:0

Velocity: 25 m/s

Time :10 sec.

Acclereation?.

Use formula for acceleration,

a =  \frac{v}{t}

a =  \frac{25}{10}

a = 2.5

Divide units to

\frac{ \frac{m}{s} }{s}  =  \frac{m}{ {s}^{2} }

So our acceleration is

2.5 \frac{m}{ {s}^{2} }

If you want to, you can use kinematic equation

v_{i} + at = v

vi is inital velocity, since v1=0 we just have

at = v

a =  \frac{v}{t}

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2 years ago
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When air pressure is lower what happens to a balloon filled with air
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3 years ago
When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f
Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer (r), measured in kilometers, can be represented by a right triangle:

r = \sqrt{x^{2}+y^{2}} (1)

Where:

x - Horizontal distance between the rocket and the observer, measured in kilometers.

y - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket (\theta), measured in sexagesimal degrees, is defined by the following trigonometric relation:

\tan \theta = \frac{y}{x} (2)

If we know that x = 5\,km, then the expression is:

\tan \theta = \frac{y}{5}

And the rate of change of this angle is determined by derivatives:

\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y

\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}

\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}

\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}

Where:

\dot \theta - Rate of change of the angle of elevation, measured in sexagesimal degrees.

\dot y - Vertical speed of the rocket, measured in kilometers per hour.

If we know that y = 4\,km and \dot y = 400\,\frac{km}{h}, then the rate of change of the angle of elevation is:

\dot \theta = 48.780\,\frac{\circ}{s}

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0
3 years ago
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