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Serjik [45]
3 years ago
9

Which type of wave has a wave perpendicular to the disturbance

Physics
1 answer:
Orlov [11]3 years ago
3 0

Answer:

Transverse wave

Explanation:

Depending on the direction of the oscillation, there are two types of waves:

- Transverse wave: in a transverse wave, the oscillation occurs in a direction perpendicular to the direction of propagation of the wave. These types of waves are characterized by alternating crests and troughs. Examples of transverse waves are electromagnetic waves.

- Longitudinal wave: in a longitudinal wave, the oscillation occurs in a direction parallel to the direction of propagation of the wave. These types of waves are characterized by alternating regions of higher density (compressions) and lower density (rarefactions). Examples of longitudinal waves are sound waves.

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A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p
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Answer:

 13.4 x 10 raise to power -19  C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

. The uniform electric field is  E = 214 N/M

, The decrease in electrical potential energy is   d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

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A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
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Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

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Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

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