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77julia77 [94]
3 years ago
6

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:
pantera1 [17]3 years ago
6 0

Answer:

It's A

Explanation:

As the waves progress through the medium, the particles they are made of move perpendicular to the direction in which the waves move. The particles do not move with the wave. So waves transmit energy but not matter as they progress through a medium.

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aleksandr82 [10.1K]

Answer:

I have examined the different parts of blub and I assume the answer is (D)T and U

Hope this helps you

4 0
3 years ago
Can someone please answer this for me
UNO [17]

C

Because in my opinion they do bending of a light wave as it passes at an angle from one medium to another.

5 0
3 years ago
Suppose the value of one division of vernier scale is 0.5mm and the value of one main scale
gregori [183]

Answer:

-0.01 mm

Explanation:

We are given that

The value of one division of vernier scale =0.5 mm

The value of one main scale division=0.49 mm

We have to find the value of least count of the instrument in mm.

We know that

Leas count of vernier caliper=1 main scale division-1 vernier scale division

Least count of vernier caliper=0.49-0.50=-0.01 mm

Hence, the least count of the instrument=-0.01 mm

Answer: -0.01 mm

8 0
3 years ago
A mobile starts with a speed of 250m / s and begins to decelerate at a rate of 3m / s². How fast is it after 45s?
Korvikt [17]

\large{ \underline{ \underline{ \bf{ \purple{Given}}}}}

  • Speed of the mobile = 250 m/s
  • It starts decelerating at a rate of 3 m/s²
  • Time travelled = 45s

\large{ \underline{ \underline{ \bf{ \green{To \: find}}}}}

  • Velocity of mobile after 45 seconds

\large{ \underline{ \underline{ \red{ \bf{Now, \: What \: to \: do?}}}}}

We can solve the above question using the three equations of motion which are:-

  • v = u + at
  • s = ut + 1/2 at²
  • v² = u² + 2as

So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

\large{ \bf{ \underline{ \underline{ \orange{Solution:}}}}}

We are provided with,

  • u = 250 m/s
  • a = -3 m/s²
  • t = 45 s

By using 1st equation of motion,

⇛ v = u + at

⇛ v = 250 + (-3)45

⇛ v = 250 - 135 m/s

⇛ v = 115 m/s

✤ <u>Final</u><u> </u><u>velocity</u><u> </u><u>of</u><u> </u><u>mobile</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u>

<u>━━━━━━━━━━━━━━━━━━━━</u>

4 0
3 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

7 0
2 years ago
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