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3 years ago

The material through which a mechanical wave travels is

Physics

1 answer:

vivado [14]3 years ago

6 0

<h2>Answer: The Medium </h2><h2 />

<u>The medium is the main factor that differentiates a mechanical wave from an electromagnetic wave,</u> since the first can not propagate without its existence, while the second can propagate regardless of whether the medium exists or not.

In addition, it is the medium that will define, the propagation speed of the wave, according to its specific physical characteristics.

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There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the pl

kaheart [24]

Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28 × 10⁵km

Orbital period T = 28.9days

T = 28.9 × 24 × 60 × 60

= 2,496,960s

Angular momentum of the moon about the planet

L = mvr

L = mr²w

$L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s$

7 0

3 years ago

Read 2 more answers

An umbrella tends to move upward on a windy day because _____.

masha68 [24]

E. all of the above

An umbrella tends to move upward on a windy day because _<span>A. buoyancy increases with increasing wind speed </span>
<span>B. air gets trapped under the umbrella and pushes it up </span>
<span>C. the wind pushes it up </span>
<span>D. a low-pressure area is created on top of the umbrella </span>

3 0

3 years ago

Three bullets are fired simultaneously by three guns aimed toward the center of a circle where they mash into a stationary lump.

earnstyle [38]

Answer:

2,87 * $10^{-3}$

Explanation:

When the bullets meet at the center and collide, since momentum is a vectoral quantity, their momentum vectors even up and are sumof zero. Formula of momentum is P = m.v , where m is mass and v is velocity. Let’s name the first two bullets as x,y and the one which mass is unknown as z. Then calculate momentum of x and y:

Px= 5,30 * $10^{-3}$ * 301 = 1,5953 kg*m/s

Py= 5,30 * $10^{-3}$ * 301 = 1,5953 kg*m/s

The angle between x and y bullets is 120°, and we know that if the angle between two equal magnitude vectors is 120°, the magnitude of the resultant vector will be equal to first two and placed in exact middle of two vectors. So we can say total momentum of x and y (Px+Py) equals to 1,5953 kg*m/s as well (Shown in the figure).

For z bullet to equalize the total momentum of x and y bullets, it needs to have the same amount of momentum in the opposite way.

Pz = 1,5953 = m * 554

m = 2,87 * $10^{-3}$ kg

8 0

3 years ago

A body travels the first half of the total distance with velocity v and second half with v2 calculate avg velocity

LuckyWell [14K]

Answer:

v = 2 v₁ v₂ / (v₁ + v₂)

Explanation:

The body travels the first half of the distance with velocity v₁. The time it takes is:

t₁ = (d/2) / v₁

t₁ = d / (2v₁)

Similarly, the body travels the second half with velocity v₂, so the time is:

t₂ = (d/2) / v₂

t₂ = d / (2v₂)

The average velocity is the total displacement over total time:

v = d / t

v = d / (t₁ + t₂)

v = d / (d / (2v₁) + d / (2v₂))

v = d / (d/2 (1/v₁ + 1/v₂))

v = 2 / (1/v₁ + 1/v₂)

v = 2 / ((v₁ + v₂) / (v₁ v₂))

v = 2 v₁ v₂ / (v₁ + v₂)

8 0

3 years ago

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