Sure what do u need help with
Answer: D. 292,338 J
This is the correct answer :)
I have the exact same question, any chance you figured it out since you posted this?
Answer:
L = 0.635m
Explanation:
This problem involves the concept of stationary waves in pipes. For pipes closed at one end,
The frequency f = nv/4L for n = 1,3,5....n
For pipes open at both ends
f = nv/2L for n = 1,2,3,4...n
Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.
The film solution can be found in the attachment below.
Answer:
The appropriate solution is:
(a) ![\frac{1}{4}(I_o)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D%28I_o%29)
(b) ![\frac{1}{4} (u_o)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D%20%28u_o%29)
(c) ![\frac{1}{2}B_o](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DB_o)
Explanation:
According to the question, the value is:
Power of bulb,
= 60 W
Distance,
= 1.0 mm
Now,
(a)
⇒ ![\frac{I}{I_o} =\frac{r_o_2}{r_2}](https://tex.z-dn.net/?f=%5Cfrac%7BI%7D%7BI_o%7D%20%3D%5Cfrac%7Br_o_2%7D%7Br_2%7D)
On applying cross-multiplication, we get
⇒ ![I=I_o\times \frac{1_2}{2^2}](https://tex.z-dn.net/?f=I%3DI_o%5Ctimes%20%5Cfrac%7B1_2%7D%7B2%5E2%7D)
⇒ ![=I_o\times \frac{1}{4}](https://tex.z-dn.net/?f=%3DI_o%5Ctimes%20%5Cfrac%7B1%7D%7B4%7D)
⇒ ![=\frac{1}{4} (I_o)](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B4%7D%20%28I_o%29)
(b)
As we know,
⇒ ![\frac{u}{u_o} =\frac{I}{I_o}](https://tex.z-dn.net/?f=%5Cfrac%7Bu%7D%7Bu_o%7D%20%3D%5Cfrac%7BI%7D%7BI_o%7D)
By putting the values, we get
⇒ ![u=\frac{1}{4}(u_o)](https://tex.z-dn.net/?f=u%3D%5Cfrac%7B1%7D%7B4%7D%28u_o%29)
(c)
⇒ ![\frac{B^2}{B_o^2} =\frac{u}{u_o}](https://tex.z-dn.net/?f=%5Cfrac%7BB%5E2%7D%7BB_o%5E2%7D%20%3D%5Cfrac%7Bu%7D%7Bu_o%7D)
![=\frac{I}{I_o}](https://tex.z-dn.net/?f=%3D%5Cfrac%7BI%7D%7BI_o%7D)
⇒ ![B=B_o\times \sqrt{\frac{1}{4} }](https://tex.z-dn.net/?f=B%3DB_o%5Ctimes%20%5Csqrt%7B%5Cfrac%7B1%7D%7B4%7D%20%7D)
⇒ ![=\frac{1}{2}(B_o)](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%28B_o%29)