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3 years ago

Particle motion in surface waves is __________ motion.

Physics

2 answers:

Alexxandr [17]3 years ago

8 0

The particle motion in surface waves is a combination of longitudinal and transverse motion.

Further Explanation:

According to the vibration of the particle, the waves are classified into two categories:

Transverse Waves

Longitudinal waves

Transverse Waves:

The transverse waves are the waves in which the motion of the wave particle is perpendicular to the direction of motion of the waves.

Example: The electromagnetic waves, vibrations produced in a guitar string etc.

Longitudinal waves:

The longitudinal waves are the waves in which the motion of the wave particle is along the direction of propagation of the wave.

Example: Sound waves, seismic waves etc.

The waves produces on the surface are considered to be the combination of the transverse as well as the longitudinal waves because during its motion on the surface, the wave particles vibrate perpendicular to the surface as well as along the direction of propagation of the wave.

Thus, the particle motion in surface waves is a combination of longitudinal and transverse motion.

Learn More:

1. What is the threshold frequency ν0 of cesium brainly.com/question/6953278

2. Calculate the wavelength of an electron (m = 9.11 × 10-28 g) moving at 3.66 × 106 m/s brainly.com/question/1979815

3. What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays brainly.com/question/9059731

Answer details:

Grade: High School

Subject: Physics

Physics: Waves

Keywords:

Transverse, longitudinal, combination, waves, surface waves, along, perpendicular, propagation, electromagnetic, surface of water.

gulaghasi [49]3 years ago

7 0

<h3>Answer;</h3>

B.)neither longitudinal nor transverse

<h3>Explanation;</h3>

Longitudinal waves are waves in which the vibration of particles is parallel to the direction of the wave motion.
Transverse waves on the other hand are those waves in which the vibration of particles is perpendicular to the direction of the wave motion.
In surface waves particles in the medium of transmission move in a circular motion. Therefore, they are neither transverse waves nor longitudinal waves.

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AysviL [449]

Answer:

2.28

Explanation:

From mirror formula,

1/f = 1/u+1/v .......... Equation 1

Where f = focal length of the mirror, v = image distance, u = object distance.

Note: The focal length mirror is positive.

make v the subject of the equation,

v = fu/(u-f)............ Equation 2

Given: f = 2.5 cm, u = 1.4 cm

Substitute into equation 2

v = 2.5(1.4)/(1.4-2.5)

v = 3.5/-1.1

v = -3.2 cm.

Note: v is negative because it is a virtual image.

But,

Magnification = image distance/object distance

M = v/u

Where M = magnification.

Given: v = 3.2 cm, u = 1.4 cm

M = 3.2/1.4

M = 2.28.

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4 years ago

If the 50 kg objects slows down to a velocity of 1 m/s how much kinetic energy does it have?

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3 years ago

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Why won't a very bright beam of red light impart more energy to an ejected electron than a feeble beam of violet light?

bearhunter [10]

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3 years ago

At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl

Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

$s=ut+0.5at^2 \\$

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

$h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9$

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

$h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\$

Solving both equations

$600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\$

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

$h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\$

Passing of B occurs at 4108.31 height.

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3 years ago

At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?

sergey [27]

1) Initial upward acceleration: $6.0 m/s^2$

2) Mass of burned fuel: $0.10\cdot 10^4 kg$

Explanation:

There are two forces acting on the rocket at the beginning:

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$m=1.9\cdot 10^4 kg$ is the rocket's mass

$g=9.8 m/s^2$ is the acceleration of gravity

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$T=3.0\cdot 10^5 N$

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

$T-mg=ma$ (1)