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Aloiza [94]
3 years ago
13

Which answer explains why plate B is moving underneath plate A

Physics
1 answer:
blsea [12.9K]3 years ago
4 0

Answer:

A.

Explanation:

Earth is composed of different layers and one layer moves over another due to differences in the densities.

According to the physics of density, a substance having less density floats over a higher density substance. The oceanic crust has more density than the continental crust that is why continental crust float over oceanic crust.

So in the given example, plate B is moving below the plate A, it means plate B is more dense than plate A because plate B is composed of oceanic crust . <u>For example : continents float over the asthenosphere (a layer below the lithosphere).</u>

Hence, the correct answer is "A ".

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A 2.0-kg pistol fires a 1.0-g bullet with a muzzle speed of 1000 m/s. The bullet then strikes a 10-kg wooden block resting on a
Andreyy89

Answer:

1000 N

Explanation:

An impulse results in a change of momentum

FΔt = mΔv

F = 0.001 kg(1000 - 0) m/s / 0.001 s = 1000 N

4 0
3 years ago
You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be
Alex

Answer:

The amplitude  is  A =  90.2 \ m

Explanation:

From the question we are told that

    The frequency of when sound is approaching observer is   f = 392 Hz

     The frequency as the move away from observer  is  f_ a =  330 \ Hz

    The time between the pitch are t =  10 \ s

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

       T =  2 t

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

          T = \frac{2 \pi}{w}

Where  w is the angular velocity

=>   \frac{2 \pi}{w}  =  2 * 10

=>   w =  0.314 \ rad/sec

Now using Doppler Effect,

   The source of the sound is approaching the observer

The

          f = f_o (\frac{v}{v- wA} )

         392  = f_o (\frac{v}{v- wA} )

Where A is the amplitude

    So when the source is moving away from the observer

         f_a =  f_o (\frac{v}{v+ wA} )  

        330  =  f_o (\frac{v}{v+ wA} )  

Here  f_o is the fundamental frequency

Dividing the both equation  we have

           \frac{392}{330}  =  \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}

           1.1878  = \frac{v+wA}{v-wA}

         1.1878 v -  1.1878 wA = v+wA

        1.1878 v = 2.1878 wA

=>     A =  \frac{(0.1878 * (330))}{(2.1878)* (0.314)}

         A =  90.2 \ m

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Answer:

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Explanation:

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