1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Rudiy27
2 years ago
11

A bell is rung. What best describes the density of air around the bell?

Physics
2 answers:
Westkost [7]2 years ago
6 0

Answer:

d.  The air density increases and decreases repeatedly before returning to normal.

Explanation:

when sound wave travel in the air then the air molecules will move to and fro about their mean position due to which sound energy propagate into the medium.

Due to this propagation of sound through the medium the pressure in the medium will go increasing and decreasing in form of rarefaction and compression and given as

P = BAK sin(\omega t + kx + \phi)

so here since pressure is changing from maximum to minimum so here density of medium also changes. So correct answer will be

d.  The air density increases and decreases repeatedly before returning to normal.

laila [671]2 years ago
4 0
D. The air density increases and decreases repeatedly before returning to normal. At rest, before the bell is rung, the density of the air is fairly undisturbed. Once the bell is rung, oscillating sound waves will cause the surrounding air to compress and expand due to the wave crests and troughs generated by the sound. This will result in a variety of destructive and constructive interference events. Eventually, the energy generated from the bell will be lost to the environment and the sound will subside as the sound waves become smaller and enough destructive interference has occurred.
You might be interested in
A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang
olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: s_c = \omega_ct = 0.233t

For the horse: s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

s_c = s_h

0.233t = 1.57 + 0.0068t^2

0.0068t^2 - 0.233t + 1.57 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}

t= \frac{0.233\pm0.11}{0.0136}

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

6 0
3 years ago
Convert 123,453 to a scientific notation
Natalija [7]

Answer:

1.23453*10^5 is scientific way

3 0
2 years ago
Who was the first person to say that earth orbits the sun?
Whitepunk [10]
The first person to say the Earth orbited the sun was Nicolaus Copernicus
7 0
3 years ago
Read 2 more answers
Capillary waves travel what than long waves
7nadin3 [17]
Faster than. Hope this helps!!!
6 0
3 years ago
Read 2 more answers
An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a p
kotykmax [81]

Answer:

a)v=1.77\times 10^6\ m/s

b)v=3.872\times 10^6\ m/s

c)v=5.5\times 10^6\ m/s

d)v=6.7\times 10^6\ m/s

e)v=7.7\times 10^6\ m/s

Explanation:

Given that

d = 2 cm

V = 200 V

u=4\times 10^5\ m/s

We know that

F = E q

F = m a

E = V/d

So

m a =  q .V/d b            

a=\dfrac{q.V}{m.d}                 ---------1

The mass of electron

m=9.1\times 10^{-31}\ kg

The charge on electron

q=1.6\times 10^{-19}\ C

Now by putting the all values in equation 1

a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2

a=1.5\times 10^{15}\ m/s^2

We know that

v^2=u^2+2as

a)

s = 0.1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}

v=\sqrt{3.16\times 10^{12}}\ m/s

v=1.77\times 10^6\ m/s

b)

s = 0.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}

v=\sqrt{1.5\times 10^{13}}\ m/s

v=3.872\times 10^6\ m/s

c)

s = 1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}

v=\sqrt{3.06\times 10^{13}}\ m/s

v=5.5\times 10^6\ m/s

d)

s = 1.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}

v=\sqrt{4.5\times 10^{13}}\ m/s

v=6.7\times 10^6\ m/s

e)

s = 2 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}

v=\sqrt{6.06\times 10^{13}}\ m/s

v=7.7\times 10^6\ m/s

5 0
3 years ago
Read 2 more answers
Other questions:
  • What subjects combine to make the field of physical science?
    13·1 answer
  • When is an object magnetic?
    13·1 answer
  • It is observed that the number of asteroids (or meteoroids) of a given diameter is roughly inversely proportional to the square
    13·1 answer
  • If the toy car moved with a velocity of 2m/s to the south for 8s; what is the total displacement of the toy car?
    9·1 answer
  • Which of these statement describes a chemical property
    11·1 answer
  • A train moves uniform velocity of 36km/h for 15 sec.find distance
    13·1 answer
  • What does the blackboard do when white light falls on it?​
    14·1 answer
  • Estás cuentas son ingresos fijos o variables o egresos fijos o variables?
    8·1 answer
  • The force of gravity of a planet is depend upon wich factors?​
    7·2 answers
  • A girl, standing on a bridge, throws a stone vertically downward with an initial velocity of 12.0 m/s, into the river below. if
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!