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2 years ago

How are sound waves different from light (and other electromagnetic, waves?

Physics

1 answer:

Blababa [14]2 years ago

8 0

Sound waves travel faster

You might be interested in

Which measurements are equal to 321 decimeters

vlabodo [156]

You are currently converting Distance and Length units from Centimeters to Feet 321 Centimeters (cm) = 10.5315 Feet (ft) This is a hard one but see if this helps if not let me now and i can try again..

7 0

3 years ago

Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo

Elena-2011 [213]

a) $3.14 \cdot 10^{-4} s$

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

The position of the tip of the lever at time t is described by the equation:

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$ (1)

The generic equation that describes a wave is

$y(t)=A sin (\frac{2\pi}{T} t)$ (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

$\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}$

Therefore, the period is

$T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s$

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when $sin(...)=1$ , and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

$\lambda=2L$

where

$\lambda$ is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

$\lambda =2(5.00)=10.0 m$

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$

And by comparing it with the general equation of a wave:

$y(t)=A sin (\frac{2\pi}{T} t)$

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

$sin(\frac{2\pi}{T}t)=1$

which means that

$y(t)=A$

And therefore in this case,

$y=0.500 cm$

So, this is the displacement.

6 0

3 years ago

Two photographers are competing for business in town. Andrea uses only film photography and Keira uses only digital photography.

____ [38]

There is no right or wrong answer, your teacher wants you to support your own answer with points. As long as the reasons make logical sense you are fine.

I think they both have valid points. Their replies are both true, but from a buyer's perspective who would you purchase services from? You would get different answers depending on who you ask.

If you choose to go old school, obviously you get an actual photo that can be stored physically. This means it is a memory that can be preserved, and it might feel more nostalgic being able to touch the photo.

On the other hand, a digitally stored photo can be altered (photoshop), but it is forever as long as the internet still exists. A physical photo would fade with time, which doesn't happen with a digital photo.

It is definitely easier to argue that digital photography has more advantages (they do, it is why nobody uses film anymore)

Points you can consider:

Can be transferred to the other side of the world instantly

Ability to make copies and print as many photos as you want

Can be stored on cloud/devices and be like that forever

Compare them with film photography to give a more solid response.

3 0

3 years ago

A car travels around a curve. What can you say about the centripetal force if the velocity is tripled?

Lesechka [4]

Answer:

F = M a = M v^2 / R

If v is increased by three the force will be increased by nine,

C) is correct

4 0

2 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

melisa1 [442]

Answer:

$v_{2.6b}=11.18\ m.s^{-1}$

$v_{7.2b}=14.19\ m.s^{-1}$

$s_{bb}=226.3305\ m$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

$t_b=21.3956\ s$

$a_y=1.1065\ m.s^{-2}$

Explanation:

Given:

initial speed of blue car, $u_b=0\ m.s^{-1}$

initial speed of yellow car, $u_y=0\ m.s^{-1}$

acceleration rate of blue car, $a_b=4.3\ m.s^{-2}$

time for which the blue car accelerates, $t_{ab}=3.3\ s$

time for which the blue car moves with uniform speed before decelerating, $t_{ub}=14.3\ s$

total distance covered by the blue car before coming to rest, $s_b=253.26 \ m$

distance at which the the yellow car intercepts the blue car just as the blue car come to rest, $s_y=253.26 \ m$

Speed of blue car after 2.6 seconds of starting the motion:

Applying the equation of motion:

$v_{2.6b}=u_b+a_b.t$

$v_{2.6b}=0+4.3\times 2.6$

$v_{2.6b}=11.18\ m.s^{-1}$

Speed of blue car after 7.2 seconds of starting the motion:

∵The car accelerates uniformly for 3.3 seconds after which its speed becomes uniform for the next 14.3 second before it applies the brake.

so,

$v_{7.2b}=u+a_b\times t_{ab}$

$v_{7.2b}=0+4.3\times 3.3$

$v_{7.2b}=14.19\ m.s^{-1}$

Distance travelled by the blue car before application of brakes:

This distance will be $s_{bb}=$ (distance travelled during the accelerated motion) + (distance travelled at uniform motion)

Now the distance travelled during the accelerated motion:

$s_{ab}=u_b.t_{ab}+\frac{1}{2} a_{b}.t_{ab}^2$

$s_{ab}=0\times 3.3+0.5\times 4.3\times 3.3^2$

$s_{ab}=23.4135\ m$

Now the distance travelled at uniform motion:

$s_{ub}=14.19\times 14.3$

$s_{ub}=202.917\ m$

Finally:

$s_{bb}=s_{ab}+s_{ub}$

$s_{bb}=23.4135+202.917$

$s_{bb}=226.3305\ m$

Acceleration of the blue car once the brakes are applied

Here we have:

initial velocity, $u=14.19\ m.s^{-1}$

final velocity, $v=0\ m.s^{-1}$

distance covered while deceleration, $s_{db}=s_b-s_{bb}$

$\Rightarrow s_{db}=253.26 -226.3305=26.9295\ m$

Using the equation of motion:

$v^2=u^2+2a_{db}.s_{db}$

$0^2=14.19^2+2\times a_{db}\times 26.9295$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

The total time for which the blue car moves:

$t_b=t_a+t_{ub}+t_{db}$ ........................(1)

Now the time taken to stop the blue car after application of brakes:

Using the eq. of motion:

$v=u+a_{db}.t_{db}$

$0=14.19-3.7386\times t_{db}$

$t_{db}=3.7956\ s$

Putting respective values in eq. (1)

$t_b=3.3+14.3+3.7956$

$t_b=21.3956\ s$

For the acceleration of the yellow car:

We apply the law of motion:

$s_y=u_y.t_y+\frac{1}{2} a_y.t_y^2$

Here the time taken by the yellow car is same for the same distance as it intercepts just before the stopping of blue car.

Now,

$253.26=0\times 21.3956+0.5\times a_y\times 21.3956^2$

$a_y=1.1065\ m.s^{-2}$

7 0

3 years ago