a standing wave in a pipe with both ends open has a frequency of 348. hz. the next higher harmonic has a frequency of 522. hz.
1 answer:
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Answer:
A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same
B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same
Explanation:
A vehicle crash problem must be solved with the equation of the moment,
Initial instant Before crash
p₀ = m v₁ + mv₂
After the crash
= m
+ m
p₀ =
If the speed ratio before and after the crash is one
p₀ / = 1
We can assume that initially one of the cars was stopped
m v₁₀ = m
v₁₀ =
For the two speeds to be equal, the masses of the vehicles must be the same.
A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same
B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same
Answer:
The answer to the question is;
The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.
Explanation:
We note that the distance traveled by each car after 4 seconds is
Car A = 19 ft in the west direction.
Car B = 26 ft in the north direction
The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the west being the x coordinate.
Therefore, let the distance between the two cars be s
we have
s² = x² + y²
= (19 ft)² + (26 ft)² = 1037 ft²
s = = 32.202 ft.
The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;
Since s² = x² + y² we have
→ which gives
We note that the speeds of the cars were given as
Car B moving north = 12 ft/sec, which is the y direction and
Car A moving west = 8 ft/sec which is the x direction.
Therefore
= 12 ft/sec and
= 8 ft/sec
becomes
= 464 ft²/sec
= 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.