How does mass effect my room?

Two loudspeakers on a concert stage are vibrating in phase. A listener 50.5 m from the left speaker and 26.0 m from the right on

mixer [17]

To solve this problem, it is necessary to apply the concepts related to the constructive interference caused by the wavelengths of the sound traveling in the air.

From the definition of constructive interference we know that

$\Delta x = n \lambda \rightarrow \lambda = \frac{\Delta x}{n}$

Where

$\Delta x =$ Distance between speakers

n = Integer which represent he number of repetition of the spectrum

$\lambda =$ Wavelength

At the same time the frequency is subject to the form,

$f = \frac{v}{\lambda}$

Where

$v = 343m/s \rightarrow$ Velocity (of the sound at this case)

From our given values we have to $\Delta x$ is

$\Delta x = 50.5m-26m$

$\Delta x = 24.5m$

The wavelength would be subject to the sound spectrum therefore for n = 1,

$\lambda = \frac{\Delta x}{n}$

$\lambda = \frac{24.5}{1}$

$\lambda = 24.5m$

Then the frequency would be,

$f = \frac{v}{\lambda}$

$f = \frac{343}{24.5}$

$f = 14Hz$

For the value of n = 2,

$\lambda = \frac{\Delta x}{n}$

$\lambda = \frac{24.5}{2}$

$\lambda = 12.25m$

Then the frequency would be,

$f = \frac{v}{\lambda}$

$f = \frac{343}{12.25}$

$f = 28Hz$

For n = 3,

$\lambda = \frac{\Delta x}{n}$

$\lambda = \frac{24.5}{3}$

$\lambda = 8.167m$

Then the frequency would be,

$f = \frac{v}{\lambda}$

$f = \frac{343}{12.25}$

$f = 42Hz$

From the frequencies obtained we can identify that the two lowest frequencies that can be heard due to constructive interference are 28Hz and 42Hz

5 0

3 years ago

Which one of the following statements is false?

emmasim [6.3K]

Answer:

d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero

Explanation:

Affirmations

a) true. The orbits are accurate in the Bohr model and probabilistic in quantum mechanics

b) True. If both give the same results and use the same quantum number (n)

c) True. If in angular momentum it is quantized, in the Bohr model too but it does not justify it

d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero

7 0

4 years ago

Which of these describes Kepler’s third law of orbital motion?(1 point)

Nadusha1986 [10]

Answer:

T2 ∝ a3

Explanation: took the quick check

4 0

2 years ago

If the rotation of a planet of radius 7.66 x 10^6 m and free-fall acceleration 8.34 m/s^2 increased to the point that the centri

madreJ [45]

Answer:

On the surface of the moon the weight of an object points towards the center of the moon and its magnitude is approximately 1/6 the magnitude of its weight on the surface of the earth. The mass of the object, i.e. its resistance to acceleration, is the same everywhere

5 0

3 years ago

In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s

nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

$W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where:

$F=1.1 \cdot 10^5 N$ is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

$d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m$