The frequencies of light that an atom can emit are dependent on states the electrons can be in. When excited, an electron moves to a higher energy level or orbital. When the electron falls back to its ground level the light is emitted.
hope this helped:)
mark brainliest
The Moon is 3.8 108 m from Earth and has a mass of 7.34 1022 kg. 5.97 1024 kg is the mass of the Earth.
<h3>What kind of gravitational pull does the moon have on the planet?</h3>
On the surface of the Moon, the acceleration caused by gravity around 1.625 m/s2 which is 16.6% greater than on the surface of the Earth 0.166.
<h3>What does the Earth's center's gravitational pull feel like?</h3>
Gravity is zero if you are in the centre of the earth since everything around you is pulling "up" (up is the only direction).
<h3>Where is the Earth's and the moon's gravitational centre?</h3>
It is around 1700 kilometres below Earth's surface.
To know more about gravitational force visit:-
brainly.com/question/12528243
#SPJ4
Answer:
500 m
Explanation:
t = Time taken
u = Initial velocity = 50 m/s
v = Final velocity = 0
s = Displacement
a = Acceleration = -2.5 m/s²
Equation of motion

Time taken by the train to stop is 20 seconds

∴ The engineer applied the brakes 500 m from the station
Answer:
the frequency of the second harmonic of the pipe is 425 Hz
Explanation:
Given;
length of the open pipe, L = 0.8 m
velocity of sound, v = 340 m/s
The wavelength of the second harmonic is calculated as follows;
L = A ---> N + N--->N + N--->A
where;
L is the length of the pipe in the second harmonic
A represents antinode of the wave
N represents the node of the wave

The frequency is calculated as follows;

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.