**Your
Answer to Q520.**
Sorry, your answer is **not** correct. Don't forget to incude the use of
the reference level *I*_{min}.

Help: *Fundamentals of
Sound*, Sec. 11-F.

Or would you like a **HINT**?

You should try to work out the answer on your own, but if you insist on reading it the correct answer is here

Return to Question 520.

**Hint
for Question 520:** To convert to dB, you divide *I* by the reference
level, 10^{-12}W/m^{2} and then find the log of that number.
The dB level is 10 times that log.

Return to Question 520.

**Your
Answer to Q520. **Congratulations, your answer is **correct**.

If you would like, you can compare your answer to the "official" correct answer

Return to Question 520.

**Correct
Answer to Question 520:** *I*/*I*_{min} = 5x10^{-8}/10^{-12}
= 5x10^{4} = 10^{4.7}. The exponent 4.7 is the log of the previous
number. The decibel level is then 47 dB, that is answer c).

Return to Question 520.

**Your
Answer to Q530.** Sorry, your answer is **not** correct. The
89 is related to the logarithm of the number you want. So you have to take an
antilog.

Help: *Fundamentals of
Sound*, Sec. 11-H.

Or would you like a **HINT**?

You should try to work out the answer on your own, but if you insist on reading it the correct answer is here

Return to Question 530.

**Hint
for Question 530:** To convert to intensity, you divide the decibel level
by 10 to convert to Bels. This number is the power of ten (log) of the ratio
of intensities *I*/*I*_{min}. Find its antilog and multiply
by the reference level *I*_{min} = 10^{-12}W/m^{2}.

Return to Question 530.

**Your
Answer to Q530. **Congratulations, your answer is **correct**.

If you would like, you can compare your answer to the "official" correct answer

Return to Question 530.

**Correct
Answer to Question 530:** 89 dB = 8.9 Bels. Thus *I*/*I*_{min}
= 10^{8.9} = 10^{0.9} x 10^{8} = 7.9 x 10^{8}.
In the last step we used the log table to find the antilog of 0.9. We multiply
the result by the reference level to get the intensity: *I* = 7.9 x 10^{8}
x 10^{-12} W/m^{2} = 7.9 x 10^{-4} W/m^{2},
which is answer b).

Return to Question 530.

**Your
Answer to Q540.** Sorry, your answer is **not** correct. Multiply
the intensity by 5 and convert normally to dB.

Help: *Fundamentals of
Sound*, Sec. 11-K.

Or would you like a **HINT**?

You should try to work out the answer on your own, but if you insist on reading it the correct answer is here

Return to Question 540.

**Hint
for Question 540:** Multiply the intensity by 5 and convert normally
to dB. To convert to dB, you divide the total intensity by the reference level,
10^{-12}W/m^{2} and then find the log of that number. The dB
level is 10 times that log.

Return to Question 540.

**Your
Answer to Q540. **Congratulations, your answer is **correct**.

If you would like, you can compare your answer to the "official" correct answer

Return to Question 540.

**Correct
Answer to Question 540:** 5 *I*/*I*_{min} = 5x10^{-2}/10^{-12}
= 5x10^{10} = 10^{10.7}. The decibel level is then 107 dB, that
is answer c).

Return to Question 540.

**Your
Answer to Q545.** Sorry, your answer is **not** correct. The
40 dB sound is quite a bit more intense than the 30 dB sound. Note that you
cannot just add the dB levels.

Help: *Fundamentals of
Sound*, Sec. 11-K.

Or would you like a **HINT**?

Return to Question 545.

**Hint
for Question 545:** The 40 dB sound is quite a bit more intense than
the 30 dB sound. Thus you know your answer is pretty close to 40 dB without
doing any math. But to determine which of the remaining possible answers it
is you may need to do the math. Incidentally you know the answer cannot be 44
dB because doubling the sound intensity raises the level by only 3 dB so that
40 dB + 40 dB is a total intensity of only 43 dB. So 44 dB is too much in this
problem.

Return to Question 545.

**Your
Answer to Q545. **Congratulations, your answer is **correct**.

If you would like, you can compare your answer to the "official" correct answer

Return to Question 545.

**Correct
Answer to Question 545:** The ratio of total intensity to reference
level is (*I*_{1} + *I*_{2} )/*I*_{min}
= 10^{4.0} + 10^{3.0} = 1 x 10^{4.0} + 0.1 x 10^{4.0}
= 1.1 x 10^{4.0} = 10^{0.04} x 10^{4.0} = 10^{4.04}.
The exact answer is then 40.4 dB and c) is closest.

Return to Question 545.

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