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3 years ago

Audible limit of the sound wave

Physics

2 answers:

Sergio039 [100]3 years ago

6 0

Answer:

20 hz - 20khz is the audible sound wave for humans

o-na [289]3 years ago

3 0

Almost 18Hz is an audible limit of the sound wave.

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At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial

elena55 [62]

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution : Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}$

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm

6 0

3 years ago

The plane flight covers 400 miles in the first hour. As you approach the landing site, it takes you an hour to cover the last 20

9966 [12]

Miles per hour is velocity so I agree with you. Without seeing any associated question, though, it's hard to say. If the question is, find the average velocity, then velocity is correct. But since change in velocity is negative, that is also negative acceleration.

3 0

4 years ago

Read 2 more answers

What would happen if the Earth stopped rotating?

trapecia [35]

Answer:

the atmosphere would still be in motion with the Earth's original 1100 mile per hour rotation speed at the equator. ... This means rocks, topsoil, trees, buildings, your pet dog, and so on, would be swept away into the atmosphere.

Explanation:

to me, this means we would proabaly be sucked into outer space and could die if no astronaut gear is on

4 0

3 years ago

The tub of a washing machine goes into its spin cycle, starting from rest and gaining angular speed steadily for 10.0 s, at whic

ioda

Answer:

69 revolutions

Explanation:

given,

initial angular velocity = 0

final angular velocity = 6 rev/s

when safety switch turned off

tub smoothly slows to rest in (t)= 13 s

total revolution of tub = ?

angular acceleration = $\dfrac{\omega_f-\omega_0}{t}$

angular acceleration = $\dfrac{6-0}{10}$

= 0.6 rev/s²

now,

$\theta = \omega_i t +\dfrac{1}{2}\alpha t^2$

$\theta = 0+\dfrac{1}{2}\times 0.6 \times 10^2$

= 30 revolution

now,

when safety switch turn off

angular acceleration = $\dfrac{\omega_f-\omega_0}{t}$

angular acceleration = $\dfrac{0-6}{13}$

= -0.462 rev/s²

now,

$\theta = \omega_i t +\dfrac{1}{2}\alpha t^2$

$\theta = 6 \times 13 -\dfrac{1}{2}\times 0.462 \times 13^2$

= 39 revolution

total revolution = 30 + 39 = 69 revolutions.

tub will rotate 69 revolutions

4 0

3 years ago

Consider a coaxial cable (like the kind that is used to carry a signal to your TV). In this cable, a current I runs in one direc

Ira Lisetskai [31]

Answer:

0 < r < r_exterior B_total = $\frac{\mu_o I}{2\pi r}$

r > r_exterior B_total = 0

Explanation:

The magnetic field created by the wire can be found using Ampere's law

∫ B. ds = μ₀ I

bold indicates vectors and the current is inside the selected path

outside the inner cable

B₁ (2π r) = μ₀ I

B₁ = $\frac{\mu_o I}{2\pi r}$

the direction of this field is found by placing the thumb in the direction of the current and the other fingers closed the direction of the magnetic field which is circular in this case.

For the outer shell

for the case r> r_exterior

B₂ = \frac{\mu_o I}{2\pi r}

This current is in the opposite direction to the current in wire 1, so the magnetic field has a rotation in the opposite direction

for the case r <r_exterior

in this case all the current is outside the point of interest, consequently not as there is no internal current, the field produced is zero

B₂ = 0

Now we can find the field created by each part

0 < r < r_exterior

B_total = B₁

B_total = $\frac{\mu_o I}{2\pi r}$

r > r_exterior

B_total = B₁ -B₂

B_total = 0

6 0

3 years ago