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oksano4ka [1.4K]
3 years ago
5

A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves

. What does this demonstrate about the propagation of waves through a medium?
A) Waves transmit energy but not matter as they progress through a medium.
B) Waves transmit matter but not energy as they progress through a medium.
C) Waves do not transmit matter or energy as they progress through a medium.
D) Waves transmit energy as well as matter as they progress through a medium.
Physics
1 answer:
KIM [24]3 years ago
4 0

Answer:A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

Explanation:

A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

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<u>Explanation:</u>

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So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

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                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

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