First overtone of open organ pipe is given as

first overtone of closed organ pipe is given as

now they are in unison so we will have




so end correction of both pipes is e = 1 cm
Answer:
ee that the lens with the shortest focal length has a smaller object
Explanation:
For this exercise we use the constructor equation or Gaussian equation
where f is the focal length, p and q are the distance to the object and the image respectively.
Magnification a lens system is
m =
= -
h ’= -\frac{h q}{p}
In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞
Let's calculate the distance to the image for each lens
f = 6.0 cm

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf
q = f = 6.0 cm
for the lens of f = 12.0 cm q = 12.0 cn
to find the size of the image we use
h ’= h q / p
where p has a high value and is the same for all systems
h ’= h / p q
Thus
f = 6 cm h ’= fo 6 cm
f = 12 cm h ’= fo 12 cm
therefore we see that the lens with the shortest focal length has a smaller object
Acceleration = velocity/ time
Acceleration = 0.7-0.3 /30= 0.01 m/s^2
Notice that velocity is calculated the final speed minus the initial !
Answer:
i cant even see that no one can i will answer it if u can make it bigger
Explanation: eyes