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labwork [276]
3 years ago
9

Match each characteristic with the type of wave it corresponds to. Longitudinal waves Transverse waves The direction the wave tr

avels is parallel to the displacement of the medium. Electromagnetic waves are an example of this type of wave. Sound waves are an example of this type of wave. If the wave propagates in the x‑direction, the medium is also disturbed in the x‑direction. If the wave propagates in the x‑direction, the medium is disturbed in the y‑ and/or z‑direction. The direction the wave travels is perpendicular to the displacement of the medium.
Physics
2 answers:
Mandarinka [93]3 years ago
4 0

Answer:

The properties are matched with the waves below;

Longitudinal :

  • The direction the wave travels is parallel to the displacement of the medium.
  • Sound waves are an example of this type of wave.
  • If the wave propagates in the x‑direction, the medium is also disturbed in the x‑direction

Transverse waves:

  • Electromagnetic waves are an example of this type of wave.
  • If the wave propagates in the x‑direction, the medium is disturbed in the y‑ and/or z‑direction.
  • The direction the wave travels is perpendicular to the displacement of the medium.

Explanation:

Longitudinal waves: These are waves in which the direction of propagation of waves is same as the direction of displacement of the medium. Examples are sound waves, seismic P waves and ultasound waves.

Transverse waves: These are waves in which the direction of propagation of waves is perpendicular to the direction of particle vibration. Examples are vibration on a spring, water waves, light waves.

zubka84 [21]3 years ago
3 0

Answer:The direction the wave travels is parallel to the displacement of the medium.LW.

Electromagnetic waves are an example of this type of wave. TW

Sound waves are an example of this type of wave.LW

Ir the wave propagates in the x‑direction, the medium is also disturbed in the x‑direction. LW

If the wave propagates in the x‑direction, the medium is disturbed in the y‑ and/or z‑direction. TW

The direction the wave travels is perpendicular to the displacement of the medium.TW

Explanación:

the main difference of Transversal AND Longitudinal waves (TW AND LW) is that the disturbation direction in the media with respect to the travel direction is well different for both cases. For TW the perturbation oscilatiew perpendicular to propagation direction of the wave.

In constrant for LW the disturbation oscilates in the same direction as the wave is propagating.

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Put all the values,

F = 20 kg × 10 m/s²

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Alex787 [66]

Answer:

The magnitude of the tangential velocity is v= 0.868 m/s

The magnitude of the resultant acceleration at that point is  a = 4.057 m/s^2

Explanation:

From the question we are told that

      The mass of the uniform disk is m_d = 40.0kg

       The radius of the uniform disk is R_d = 0.200m

       The force applied on the disk is F_d = 30.0N

Generally the angular speed i mathematically represented as

             w = \sqrt{2 \alpha  \theta}

Where \theta is the angular displacement given from the question as

           \theta  = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}

                 =1.257\  rad

   \alpha is the angular acceleration which is mathematically represented as

                    \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

    The moment of inertial is mathematically represented as

                     I = \frac{1}{2} m_dR^2_d

Substituting values

                    I = 0.5 * 40 * 0.200^2

                        = 0.8kg \cdot m^2

Considering the equation for angular acceleration

               \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

Substituting values

               \alph\alpha = \frac{(30.0)(0.200)}{0.8}

                   = 7.5 rad/s^2

Considering the equation for angular velocity

    w = \sqrt{2 \alpha  \theta}

Substituting values

     w =\sqrt{2 * (7.5) * 1.257}

         = 4.34 \ rad/s

The tangential velocity of a given point on the rim is mathematically represented as

                 v = R_d w

Substituting values

                    = (0.200)(4.34)

                     v= 0.868 m/s

The radial acceleration at hat point  is mathematically represented as

            \alpha_r = \frac{v^2}{R}

                  = \frac{0.868^2}{0.200^2}

                 = 3.7699 \ m/s^2

The tangential acceleration at that point is mathematically represented as

               \alpha _t = R \alpha

Substituting values

           \alpha _t = (0.200) (7.5)

                 = 1.5 m/s^2

The magnitude of resultant acceleration at that point is

                 a = \sqrt{\alpha_r ^2+ \alpha_t^2 }

Substituting values

                a = \sqrt{(3.7699)^2 + (1.5)^2}

                   a = 4.057 m/s^2

         

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Answer:

D

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