Ball thrown into the air at an angle.
Answer:
B) The cosmic background radiation is expected to contain spectral lines of hydrogen and helium, and it does.
Explanation:
Answer:
Angular velocity is same as frequency of oscillation in this case.
ω =
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Explanation:
- write the equation F(r) = -K
with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω =
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Answer:
the angle of incidence θ is 45.56 º
Explanation:
Given data
strikes the mirror before wall x = 30.7 cm
reflected ray strikes the wall y = 30.1 cm
to find out
the angle of incidence θ
solution
let us consider ray is strike at angle θ so after strike on surface ray strike to wall at angle 90 - θ
we will apply here right angle triangle rule that is
tan( 90 - θ) = y /x
tan( 90 - θ) = 30.1 / 30.7
90 - θ = tan^-1 (30.1/30.7)
90 - θ = 44.4345
θ = 45.56 º
the angle of incidence θ is 45.56 º
Answer:
It is an SI unit
Explanation:
The metre is defined as the length of the path travelled by light in a vacuum in 1299 792 458 of a second. The metre was originally defined in 1793 as one ten-millionth of the distance from the equator to the North Pole