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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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Example
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I believe I seen on google if you go to Mather
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3 years ago
The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.17 Hz, and the acc
aksik [14]

Answer:

d = 8.4 cm

Explanation:

In order to calculate the amplitude of oscillation of the top of the building, you use the following formula for the max acceleration of as simple harmonic motion:

a_{max}=A\omega^2           (1)

A: amplitude of the oscillation

w: angular speed of the oscillation = 2\pif

f: frequency = 0.17Hz

The maximum acceleration of the top of the building is a 2.0% of the free-fall acceleration. Then, you have:

a_{max}=0.02(9.8m/s^2)=0.196\frac{m}{s^2}

Then, you solve for A in the equation (1) and replace the values of the parameters:

A=\frac{a_{max}}{\omega^2}=\frac{a_{max}}{4\p^2i f^2}\\\\A=\frac{0.196m/s^2}{16\pi^2(0.17Hz)^2}\\\\A=0.042m=4.2cm

The total distance, side to side, of the oscilation of the top of the building is twice the amplitude A. Then you obtain:

d = 2A = 2(4.2cm) = 8.4cm

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