<span>hair follicle
should be your answer</span>
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed

, and an accelerated motion on the y-axis, with initial speed

and acceleration

:


where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).
To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring

Therefore:

which has two solutions:

is the time of the beginning of the motion,

is the time at which the projectile hits the ground.
Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
Answer:
(a) 4.38 s.
(b) 1.817 s
Explanation:
(a)
Using
θ = ω₀t +1/2αt² ................ Equation 1
Where θ = number of revolution, t = time, α = angular acceleration, ω₀ = angular velocity.
Given: θ = 1.59 rev = 1.59×2π = 9.992 rad, ω₀ = 0 rad/s, α = 1.04 rad/s².
Substitute into equation 1
9.992 = 0(t) + 1/2(1.04)(t²)
t² = (2×9.992)/1.04
t² = 19.984/1.04
t = √(19.215)
t =4.38 s.
(b)
also using
θ = ω₀t +1/2αt²............... Equation 1
Given: θ =3.18 rev = 3.18×2π = 19.97 rad, ω₀ = 0 rad/s, α = 1.04 rad/s².
Substitute into equation 1
19.97 = 0(t) + 1/2(1.04)(t²)
t² = 19.97×2/1.04
t = √(38.40)
t = 6.197 s
The time require = 6.197-4.38 = 1.817 s
The speed of sound at

is approximately v=343 m/s. The distance covered by the sound wave is

And the time it takes is

Now we want to find how far the light travels during this time. Light travels at speed

, therefore the distance it covers during this time is
Answer:
c. testing student opinions
Explanation:
opinions aren't factual and they would not aide an experiment if it wasn't for a social experiment.