The answer is orbit, we are orbiting the sun as the moon orbits us
Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .
Answer:
<em>The 6000 lines per cm grating, will produces the greater dispersion .</em>
Explanation:
A diffraction grating is an optical component with a periodic (usually one that has ridges or rulings on their surface rather than dark lines) structure that splits and diffracts light into several beams travelling in different directions.
The directions of the light beam produced from a diffraction grating depend on the spacing of the grating, and also on the wavelength of the light.
For a plane diffraction grating, the angular positions of principle maxima is given by
(a + b) sin ∅n = nλ
where
a+b is the distance between two consecutive slits
n is the order of principal maxima
λ is the wavelength of the light
From the equation, we can see that without sin ∅ exceeding 1, increasing the number of lines per cm will lead to a decrease between the spacing between consecutive slits.
In this case, light of the same wavelength is used. If λ and n is held constant, then we'll see that reducing the distance between two consecutive slits (a + b) will lead to an increase in the angle of dispersion sin ∅. So long as the limit of sin ∅ not greater that one is maintained.
Answer:
20m/s
Explanation:
it covers 20 metres in a second