Answer:
2270 full wavelengths.
Explanation:
The wavelength 
of the sound wave (assuming sound speed of 
) is
.
Now, assuming the distance to the last row is 1947m, the number 
of wavelengths that fit into this distance are
which is 2270 full wavelengths.
Hence, there are 2270 full wavelengths between the stage and the last row of the crowd.
Answer:
6.0 s
98 m/s
Explanation:
The radius of the planet is much bigger than the height of the tower, so we will assume the acceleration is constant. Neglect air resistance.
Acceleration due to gravity on this planet is:
a = GM / r²
a = (6.67×10⁻¹¹ m³/kg/s²) (2.7 × 1.48×10²³ kg) / (1.7 × 750,000 m)²
a = 16.4 m/s²
The height of the tower is:
Δy = 96 × 3.05 m
Δy = 293 m
Given v₀ = 0 m/s, find t and v.
Δy = v₀ t + ½ at²
(293 m) = (0 m/s) t + ½ (16.4 m/s²) t²
t = 6.0 s
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (16.4 m/s²) (293 m)
v = 98 m/s
Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)
