Look at the picture and help me plz
2 answers:
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Answer:
Explanation:
Given
height of wall=5.15 m
angle of launch
Launch velocity(u)=52.4 m/s
Time of flight will be sum of time of flight of projectile+time to cover 5.15 m
Time of flight of arrow
Now time require to cover 5.15 m
Here at the time of zero vertical displacement of arrow i.e. when arrow is at the same height as of building then its vertical velocity will change its sign compared to initial vertical velocity.
at zero vertical displacement
Thus time required will be
total time =
(b)Horizontal distance=Range of arrow(R_1) + horizontal distance in 0.118 s
=263.28+3.546=266.82 m
For a pipe closed at one end, the second overtone is the 3rd natural frequency as shown in the diagram.
For the fundamental frequency,
λ = 4L
where
λ = wavelength
L = the length of the tube
For the second overtone (3rd frequency)
λ = (4/3)L
Because L = 6.8 m, therefore
λ = (4/3)*(6.8 m) = 9.0667 m
The distance between a node and an adjacent antinode is λ/2, which is
9.0667/2 = 4.533 m
Answer: 4.5 m
For the fundamental frequency,
λ = 4L
where
λ = wavelength
L = the length of the tube
For the second overtone (3rd frequency)
λ = (4/3)L
Because L = 6.8 m, therefore
λ = (4/3)*(6.8 m) = 9.0667 m
The distance between a node and an adjacent antinode is λ/2, which is
9.0667/2 = 4.533 m
Answer: 4.5 m