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3 years ago

As you move farther away from a source emitting a pure tone, the ___________ of the sound you hear decreases.

Physics

1 answer:

Colt1911 [192]3 years ago

7 0

Answer:

frequency

Explanation:

The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.

So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is

$f' = \frac{v + v_o}{v+ v_s} f$

where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.

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En la siguiente expresión matemáticas w=mg el peso w con relación a se relaciona con la masa m en una proporción

s2008m [1.1K]

Answer:

a) Directamente proporcional

Explanation:

El peso se puede definir como la fuerza que actúa sobre un cuerpo o un objeto como resultado de la gravedad.

Matemáticamente, el peso de un objeto viene dado por la fórmula;

$Peso = mg$

Donde;

m es la masa del objeto.

g es la aceleración debida a la gravedad.

De la expresión matemática, podemos deducir que el valor del peso de un objeto es directamente proporcional a la masa del objeto.

Por lo tanto, un aumento en la masa de un objeto provocaría un aumento en el peso del objeto y viceversa.

4 0

3 years ago

Two states of matter are described below. State A: Cannot be compressed and retains its shape State B: Highly compressible Which

telo118 [61]

Answer:

State A = piece of metal; State B = air

Explanation:

For the three main states of matter here's how it breaks down.

Solid - Cannot be compressed and retains its shape

Liquid - Cannot be compressed and does not retain its shape

Gas - Compressible and does not retain its shape.

Knowing this State A has to be solid. Only one of the options has A as a solid, so that's the answer. Worth knowing state B is a gas though, only one compressible, just like solid is the only one that retains its shape.

7 0

3 years ago

Read 2 more answers

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago

A space traveler discovers that her weight on a new planet is 192 newtons. Her mass is 68 kilograms. What is the gravitational a

alina1380 [7]

Answer:

g ≈ 2.82 m/s^2

Explanation:

By W = mg,

W = weight (in newtons)

m = mass (in kg)

g = gravitational acceleration (in m/s^2)

$192 = 68g$

$g = 2.82352941176 m/s^2$

g ≈ 2.82 m/s^2

3 0

2 years ago

When cs-137 decays, it emits gamma radiation. the energy of one photon is 1.06 × 10-13 j. what is the wavelength of this radiati

Kryger [21]

To find the solution to the problem, we would be using Planck's equation which is E = hv
Where:
E = energy
h = Planck's constant = 6.626 x 10-34 J·s
ν = frequency
Then, you’ll need a second equation which is c = λν
Where:
c = speed of light = 3 x 108 m/sec
λ = wavelength
ν = frequency
Reorder the equation to solve for frequency:ν = c/λ
Next, substitute frequency in the first equation with c/λ to get a formula you can use:
E = hν
E = hc/λ
But we are looking for the wavelength, so rearrange it more, then our final equation would be:
λ = hc / E
λ = (6.625E-34)(3.0E8 m/s) / (1.06E-13)
λ = 1.875E-12 m

3 0

3 years ago