Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
The statement "<span>The motion of a pendulum for which the maximum displacement from equilibrium does not change is an example of simple harmonic motion." is true.
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Answer
given,
diameter of sphere = 5 cm
radius of sphere = -2.5 cm
refractive index for glass n₁ = 1.5
refractive index for air n₂ = 1
magnification of the glass = ?
now,
S' = - 10 cm
magnification
m = + 3
The wavelength will remain unchanged.
Explanation:
The velocity 
of a wave in terms of its wavelength 
and frequency 
is
(1)
so if we double both the velocity and the frequency, the equation above becomes
(2)
Solving for the wavelength from Eqn(2), we get
We would have gotten the same result had we used Eqn(1) instead.
