An insulator which is also called a 'dielectric'.
Answer:
1/2 Hz
Explanation:
A simple harmonic motion has an equation in the form of
where A is the amplitude, 
is the angular frequency and 
is the initial phase.
Since our body has an equation of x = 5cos(π t + π/3) we can equate 
and solve for frequency f
f = 1/2 Hz
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 .
-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.
-- The mass of the satellite makes no difference.
Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is
(3.95 x 10⁵m) + (4.2 x 10⁶m)
= (3.95 x 10⁵m) + (42 x 10⁵m)
= 45.95 x 10⁵ m
The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.
The bird completes a revolution every 2.0 hours,
so its speed in orbit is
(91.9 π x 10⁵ m) / 2 hr
= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)
= 0.04 x 10⁵ m/sec
= 4 x 10³ m/sec
(4 kilometers per second)
