All categories

2 years ago

Which wave has the smallest amplitude?

Physics

1 answer:

nydimaria [60]2 years ago

8 0

Answer:

C. C

Explanation:

A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.

In Science, there are two (2) types of wave and these include;

I. Electromagnetic waves: it doesn't require a medium for its propagation and as such can travel through an empty space or vacuum. An example of an electromagnetic wave is light.

II. Mechanical waves: it requires a medium for its propagation and as such can't travel through an empty space or vacuum. An example of a mechanical wave is sound.

A crest can be defined as the highest (vertically) point on a waveform.

On a related note, a trough is the lowest (vertically) on a waveform.

An amplitude can be defined as a waveform that's measured from the center line (its origin or equilibrium position) to the bottom of a trough or top of a crest. Thus, the vertical axis (y-axis) is the amplitude of a waveform i.e it's measured vertically.

In this scenario, waveform C which is represented by a blue curvy line has the smallest amplitude in comparison with the other waveforms because it has the minimum height when measured from the origin.

In contrast, waveform A represented by a purple line has the highest amplitude because it has the maximum height when measured from the origin.

Mathematically, the amplitude of a wave is given by the formula;

x = Asin(ωt + ϕ)

Where;

x is displacement of the wave measured in meters.
A is the amplitude.
ω is the angular frequency measured in rad/s.
t is the time period measured in seconds.
ϕ is the phase angle.

You might be interested in

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

3 years ago

A slinky is stretched by two forces. When the forces are removed, the slinky returns to its original length. The slinky has been

e-lub [12.9K]

Answer: elastically deformed or non-permanently deformed

Explanation:

According to classical mechanics, there are two types of deformations:

-Plastic deformation (also called irreversible or permanent deformation), in which the material does not return to its original form after removing the applied force, therefore it is said that the material was permanently deformed.

This is because the material undergoes irreversible thermodynamic changes while it is subjected to the applied forces.

-Elastic deformation (also called reversible or non-permanent deformation), in which the material returns to its original shape after removing the applied force that caused the deformation.

In this case the material also undergoes thermodynamic changes, but these are reversible, causing an increase in its internal energy by transforming it into elastic potential energy.

Therefore, the situation described in the question is related to elastic deformation.

8 0

2 years ago

What distance does light travel in water, glass, and diamond during the time that it travels 1.0 m in vacuum? The refractive ind

lidiya [134]

Answer:

refractive index for water,glass,diamond are 0.752m, 0.667m, 0.413m respectively

Explanation:

refractive index (n) = $\frac{velocity of light in air/vacuum}{velocity of light in substance}$

velocity = $\frac{distance}{time}$

The time for travel is kept constant for all mediums.

refractive index (n) = $\frac{\frac{distance in vacuum}{time} }{\frac{distance in medium}{time} }\\ \\=\frac{distance in vacuum}{distance in medium}$

distance in medium = $\frac{distance in vacuum}{refractive index of medium}$

$S_{medium} = \frac{S_{vacuum} }{n_{medium} }$

For water, n= 1.33

$S_{water} = \frac{1 }{1.33}$

$S_{water} = 0.752m$

For glass, n=1.5

$S_{glass}= \frac{1 }{1.5}$

$S_{glass} = 0.667m$

For diamond, n= 2.42

$S_{diamond} = \frac{1 }{2.42}$

$S_{diamond} = 0.413m$

3 0

2 years ago

if a torque of 55.0 N/m is required and the largest force that can be exerted by you is 135 N what is th e length of the lever a

Whitepunk [10]

Answer:

$r=0.41m$

Explanation:

Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

$\tau=r\times F$

Due to the definition of cross product, the magnitude of the torque is given by:

$\tau=rFsin\theta$

Where $\theta$ is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when $sin\theta$ is equal to one, solving for r:

$r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m$

7 0

2 years ago

A baseball pitcher throws a ball at 40 ms^-1. if the acceleration is approximately constant over a distance of 2 m, how large is

Reil [10]

We have the equation of motion $v^2=u^2+2as$ , where v i the final velocity, u is the initial velocity, a is the acceleration and s is the displacement

Here final velocity, v = 40m/s

Initial velocity, u = 0 m/s

Displacement s = 2 m

Substituting $40^2=0^2+2*a*2\\ \\ a=400m/s^2$

So the baseball pitcher accelerates at 400m/ $s^2$ to release a ball at 40 m/s.

4 0

2 years ago