The Speed of Light.
Photons emitted from the surface of the sun to travel across the vacuum of space to reach out eyes
Answer:
option D
Explanation:
given,
length of the pipe, L = 0.96 m
Speed of sound,v = 345 m/s
Resonating frequency when both the end is open
![f = \dfrac{nv}{2L}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7Bnv%7D%7B2L%7D)
n is the Harmonic number
2nd overtone = 3rd harmonic
so, here n = 3
now,
![f = \dfrac{3\times 345}{2\times 0.96}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7B3%5Ctimes%20345%7D%7B2%5Ctimes%200.96%7D)
f = 540 Hz
The common resonant frequency of the string and the pipe is closest to 540 Hz.
the correct answer is option D
Answer:
The principle of momentum conservation states that if there no external force the total momentum of the system before and after the collision is conserved.
Since momentum is a vector, we should investigate the directions and magnitudes of initial and final momentum.
![\vec{P}_{initial} = \vec{P}_{final}\\\vec{P}_{initial} = m_1\vec{v}_1 + 0\\\vec{P}_{final} = m_1\vec{v}_1' + m_2 \vec{v}_2'](https://tex.z-dn.net/?f=%5Cvec%7BP%7D_%7Binitial%7D%20%3D%20%5Cvec%7BP%7D_%7Bfinal%7D%5C%5C%5Cvec%7BP%7D_%7Binitial%7D%20%3D%20m_1%5Cvec%7Bv%7D_1%20%2B%200%5C%5C%5Cvec%7BP%7D_%7Bfinal%7D%20%3D%20m_1%5Cvec%7Bv%7D_1%27%20%2B%20m_2%20%5Cvec%7Bv%7D_2%27)
If the first ball hits the second ball with an angle, we should separate the x- and y-components of the momentum (or velocity), and apply conservation of momentum separately on x- and y-directions.
![\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Cboxed%7B%20%5Cbf%7B%20%5Ccolor%7Bred%7D%7BUniversal%20%5C%3A%20law%20%5C%3A%20of%20%5C%3A%20gravitation%7D%7D%7D%7D)
Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.
❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.
According to universal law of gravitation,
![\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Clongrightarrow%7B%20%5Crm%7BF%20%5Cpropto%20m_1%20m_2%7D%7D%7D)
Also,
![\large{ \longrightarrow{ \rm{ F \propto \dfrac{1}{ {d}^{2} } }}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Clongrightarrow%7B%20%5Crm%7B%20F%20%5Cpropto%20%20%5Cdfrac%7B1%7D%7B%20%7Bd%7D%5E%7B2%7D%20%7D%20%7D%7D%7D)
Combining both, We will get
![\large{ \longrightarrow{ \rm{F \propto \dfrac{ m_1 m_2}{ {d}^{2}}}}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Clongrightarrow%7B%20%5Crm%7BF%20%20%5Cpropto%20%20%5Cdfrac%7B%20m_1%20m_2%7D%7B%20%7Bd%7D%5E%7B2%7D%7D%7D%7D%7D%20)
Or, We can write it as,
![\large{ \longrightarrow{ \rm{F \propto \: G \dfrac{ m_1 m_2}{ {d}^{2} }}}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Clongrightarrow%7B%20%5Crm%7BF%20%20%5Cpropto%20%20%5C%3A%20%20G%20%5Cdfrac%7B%20m_1%20m_2%7D%7B%20%7Bd%7D%5E%7B2%7D%20%7D%7D%7D%7D)
Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.
☯️ Hence, derived !!
<u>━━━━━━━━━━━━━━━━━━━━</u>
Answer:
vₐ = v_c
Explanation:
To calculate the escape velocity let's use the conservation of energy
starting point. On the surface of the planet
Em₀ = K + U = ½ m v_c² - G Mm / R
final point. At a very distant point
Em_f = U = - G Mm / R₂
energy is conserved
Em₀ = Em_f
½ m v_c² - G Mm / R = - G Mm / R₂
v_c² = 2 G M (1 /R - 1 /R₂)
if we consider the speed so that it reaches an infinite position R₂ = ∞
v_c =
now indicates that the mass and radius of the planet changes slightly
M ’= M + ΔM = M (
)
R ’= R + ΔR = R (
)
we substitute
vₐ =
let's use a serial expansion
√(1 ±x) = 1 ± ½ x +…
we substitute
vₐ = v_ c (
)
we make the product and keep the terms linear
vₐ = v_c