I think the correct answer would be old and metal poor stars are found in the galactic nucleus. This nucleus us a region in the center of a galaxy which contains a higher luminosity than other parts. It produces very high amounts of energy. Hope this helps.
Answer:
860.6 years.
Explanation:
The parameters given are;
Initial detector activity = 370000 alpha decays per second
Final detector activity = 93000 alpha decays per second
Formula for time to change in activity is given by the following relation;
t₉₃₀₀₀ = 2.72 × 10¹⁰ seconds = 860.6 years.
Answer:
E = hf or
Explanation:
The expression that gives the amount of light energy that is converted to other forms between the fluorescence excitation and emission events is given by :
E = hf
f is the frequency
c is the speed of light
is the wavelength
Hence, this is the required solution.
Answer:
The amplitude of the oscillating electric field is 1316.96 N/C
Explanation:
Given;
frequency of the wave, f = 2.4 Hz
intensity of the wave, I = 2300 W/m²
Amplitude of oscillating magnetic field is given by;
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
I is intensity of wave
c is speed of light = 3 x 10⁸ m/s
The amplitude of the oscillating electric field is given by;
E₀ = cB₀
E₀ = 3 x 10⁸ x 4.3899 x 10⁻⁶
E₀ = 1316.96 N/C
Therefore, the amplitude of the oscillating electric field is 1316.96 N/C
Answer:
The orbital period of the planet is 387.62 days.
Explanation:
Given that,
Mass of planet
Mass of star
Radius of the orbit
Using centripetal and gravitational force
The centripetal force is given by
We know that,
....(I)
The gravitational force is given by
....(II)
From equation (I) and (II)
Where, m = mass of planet
m' = mass of star
G = gravitational constant
r = radius of the orbit
T = time period
Put the value into the formula
Hence, The orbital period of the planet is 387.62 days.