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3 years ago

State examples of a transverse wave.

Physics

2 answers:

blondinia [14]3 years ago

7 0

Answer

Light and other types of electromagnetic radiation are examples of transverse waves. Some other examples of transverse waves include a ripple on a pond and a wave in a string.

laiz [17]3 years ago

6 0

ripples on the surface of water.

vibrations in a guitar string.

a Mexican wave in a sports stadium.

electromagnetic waves – eg light waves, microwaves, radio waves.

seismic S-waves.

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if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?

madreJ [45]

Answer:

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

$a_{c} =\frac{(velocity)^{2} }{radius}$

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

$a_{c} = ?$

Solution:

We have,

$a_{c} =\frac{(velocity)^{2} }{radius}$

Substituting given value in it we get

$a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2$

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

7 0

3 years ago

A charged particle of mass 0.0050 kg is subjected to a 5.0 T magnetic field which acts at a right angle to its motion. If the pa

Irina18 [472]

Answer:

0.01 C

Explanation:

Applying,

F = qvBsinФ................ Equation 1

Where F = Force on the charged particle, q = charge on the particle, v = velocity, B = magnetic field, Ф = angle

Since the charged particle noves in a circle,

F = mv²/r................. Equation 2

Where m = mass of the particle, v = velocity of the particle, r = radius of the circle

Substitute equation 2 into equation 1

mv²/r = qvBsinФ

make q the subject of the equation

q = mv/(rBsinФ)............. Equation 3

Given: m = 0.005 kg, v = 2 m/s, r = 0.2 m, B = 5 T, Ф = 90° (Act at right angle)

Substitute these values into equation 3

q = (0.005×2)/(0.2×5×sin90°)

q = 0.01/(1)

q = 0.01 C

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3 years ago

An object at rest will ______ unless acted upon by an unbalanced force.

zvonat [6]

Answer:

It's C because if you were trying to put it at rest that means you would put it on a Balanced surfest

4 0

2 years ago

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Oliga [24]

Answer:

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Explanation:

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3 years ago

Light in a vacuum travels at a constant speed of 3x10^8 m/s. If the moons average distance from the earth is 38776106 km how lon

Karo-lina-s [1.5K]

Answer: 258.3 s

Explanation:

The speed $s$ is given by the following equation:

$s=\frac{D}{t}$

Where:

$s=3(10)^{8} m/s$ is the speed of light in vacuum

$D=2(38776106 km \frac{1000 m}{1 km})=7.75(10)^{10} m$ is the double of the distance between Earth and Moon, since the beam of light travels from Earth to the Moon and back to Earth again.