The answer would be it will be longer than the 656.3 nm. The reduced mass of positronium is less than hydrogen so the photon energy will be a reduced amount of for positronium than for hydrogen. So this will mean that the wavelength will be lengthier.
Explanation:
We know that the number of complete waves formed in 1 sec time is frequency and the distance between two consecutive crests or troughs is wavelength. And we have the formula that
Velocity = wavelength * frequency
or, frequency = velocity / wavelength
Here we can see frequency is directly proportional to velocity and indirectly proportional to wavelength.
So as the wavelength increases frequency decreases and as the wavelength decreases frequency increases.
Hope you understood
Answer:
The correct answer is option D
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
Answer:
The magnitude of the free-fall acceleration at the orbit of the Moon is 
(
, where 
).
Explanation:
According to the Newton's Law of Gravitation, free fall acceleration (
), in meters per square second, is directly proportional to the mass of the Earth (
), in kilograms, and inversely proportional to the distance from the center of the Earth (
), in meters:
(1)
Where:
- Gravitational constant, in cubic meters per kilogram-square second.
- Mass of the Earth, in kilograms.
- Distance from the center of the Earth, in meters.
If we know that 
, 
and 
, then the free-fall acceleration at the orbit of the Moon is:
