Ask question

Ask question

All categories

3 years ago

7

Describe a compression and a rarefaction of a sound wave traveling through air

1 answer:

Nikitich [7]3 years ago

3 0

Air molecules are pushed together in a compression and air molecules are spread out in rarefraction.

You might be interested in

If two objects are the same size, will they have the same mass?

jeka57 [31]

No because they could be different materials for example one could be concrete and one could be aluminium

4 0

3 years ago

If a car is accelerating downhill under a net force of 3674 N, what force must the brakes exert to cause the car to have constan

nlexa [21]

Answer:

Explanation:

hi

5 0

3 years ago

A scientist needs to determine the average volume of five water samples collected for an experiment. What is

lorasvet [3.4K]

Answer:

D

Explanation:

ew

4 0

3 years ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

leonid [27]

Answer:

$KE_A=33\ J$

$KE_B=99\ J$

Explanation:

Given:

Let mass of the particle B be, $m_B=m$

then the mass of particle A, $m_A=3m$

Energy stored in the compressed spring, $E=132\ J$

Now when the compression of the particles with the spring is released, the spring potential energy must get converted into the kinetic energy of the particles and their momentum must be conserved.

Kinetic energy:

$\frac{1}{2}m_A.v_A^2+\frac{1}{2}m_B.v_B^2=132$

$3m.v_A^2+m.v_B^2=264$ .............................(1)

<u>Using the conservation of linear momentum:</u>

$m_A.v_A+m_B.v_B=0$

$3m.v_A+m.v_B=0$ .............................(2)

Put the value of $v_A$ from eq. (2) into eq. (1)

$3m\times (\frac{-v_B}{3})^2+m.v_B^2=264$

$v_B^2=\frac{198}{m}$ ...........................(3)

<u>Now the kinetic energy of particle B:</u>

$KE_B=\frac{1}{2}\times m_B\times v_B^2$

$KE_B=\frac{1}{2}\times m\times \frac{198}{m}$

$KE_B=99\ J$

Put the value of $v_B^2$ form eq. (3) into eq. (1):

$v_A^2=\frac{22}{m}$

<u>Now the kinetic energy of particle A:</u>

<u /> $KE_A=\frac{1}{2}m_A.v_A^2$ <u />

<u /> $KE_B=\frac{1}{2}\times 3m\times \frac{22}{m}$ <u />

$KE_A=33\ J$

6 0

4 years ago

Accurate data _____. A)is free from error B)proves a hypothesis C)disproves a hypothesis D)is always precise

maxonik [38]

The correct choice for this problem is (a)

8 0

3 years ago

Read 2 more answers