A sound wave leaves the loudspeaker. As it travels, it experiences a temporary increase in wavelength and then returns to its original wavelength. The sound wave traveled through a helium balloon (helium is less dense than air could explain this change in wavelength
The pattern of disruption brought on by energy moving away from the sound source is known as a sound wave. Longitudinal waves are what makeup sound. This indicates that the direction of energy wave propagation and particle vibrational propagation are parallel. The atoms oscillate when they are put into vibration.
A high-pressure and a low-pressure zone are created in the medium as a result of this constant back and forth action. Compressions and rarefactions, respectively, are terms used to describe these high- and low-pressure zones. The sound waves go from one medium to another as a result of these regions being transmitted to the surrounding media.
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Explanation:
It is given that,
Force, 
Position vector, 
(a) The torque on the particle about the origin is given by :
(b) To find the angle between r and F use dot product formula as :
Hence, this is the required solution.
Friction will slow down the moving object
Answer:
269 m
45 m/s
-58.6 m/s
Explanation:
Part 1
First, find the time it takes for the package to land. Take the upward direction to be positive.
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 5.98 s
Next, find the horizontal distance traveled in that time:
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²
Δx = 269 m
Part 2
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: v
v = at + v₀
v = (0 m/s²) (5.98 s) + (45 m/s
v = 45 m/s
Part 3
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)
v = -58.6 m/s
