All categories

2 months ago

Below you are given data about a wave in three different substances.

Physics

1 answer:

inna [3.1K]2 months ago

7 0

1) The wave's period remains constant across different media

2) The wave's velocity varies depending on the medium it travels through

3) As a wave transitions between media, its speed, direction, and wavelength can change, while its frequency stays unchanged

Clarification:

The period of a wave signifies the duration it takes for one full oscillation.

The wave's period is the inverse of its frequency:

$T=\frac{1}{f}$

where

T denotes the period

f is the frequency

The provided table illustrates that the frequency remains consistent across the three media; hence, the period is unchanged as it solely relies on frequency. We can compute it as we know that

f = 350 Hz

thus the period equals

$T=\frac{1}{350}=2.86\cdot 10^{-3} s = 2.86 ms$

The velocity of a wave can be derived from the wave equation:

$v=f \lambda$

where

f indicates the frequency

$\lambda$ is the wavelength

$f=350 Hz, \lambda = 0.75 m$ , resulting in a speed of

$v_1 = (350)(0.75)=262.5 m/s$

In the second medium,

$f=350 Hz, \lambda = 0.70 m$ , leading to a speed of

$v_2 = (350)(0.70)=245 m/s$

In the third medium,

$f=350 Hz, \lambda = 0.65 m$ , showing a speed of

$v_3 = (350)(0.65)=227.5 m/s$

As a result, we conclude that the wave's speed varies with the medium.

- The wave's direction alters. Specifically, if the subsequent medium is of greater optical density, the wave bends towards the normal; conversely, it bends away if the second medium is of lesser optical density.

- The wave's speed is affected. The wave decelerates in media with higher optical density and accelerates in those with lower optical density.

- The wave's frequency remains unchanged.

- Ultimately, the wave's wavelength is modified. If moving into a medium of greater optical density, the wavelength decreases, while it increases in one of lower optical density.

Discover more about waves here:

</pwhen></pin>

You might be interested in

Consult Conceptual Example 9 in preparation for this problem. Interactive LearningWare 6.3 also provides useful background. The

Keith_Richards [3271]

Answer:

11.56066 m/s

Explanation:

m = Mass of individual

v = Velocity of individual = 13.4 m/s

g = Gravitational acceleration = 9.81 m/s²

v' = Velocity of the individual after dropping

At the surface, kinetic and potential energy will equalize

$\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{13.4^2}{2\times 9.81}\\\Rightarrow h=9.15188\ m$

The cliff's height is 9.15188 m

Define fall height as h' = 2.34 m

$\dfrac{1}{2}mv'^2+mgh'=mgh\\\Rightarrow v'=\sqrt{2g(h-h')}\\\Rightarrow v'=\sqrt{2\times 9.81(9.15188-2.34)}\\\Rightarrow v'=11.56066\ m/s$

The person's speed is 11.56066 m/s

3 0

2 months ago

The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu

Yuliya22 [3333]

Answer: $0.10233nm$

Explanation:

The mean free path of an atom can be calculated using the following equation:

$\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}$ (1)

Where:

$\lambda=0.2\mu m=0.2(10)^{-6}m$

$R=8.3145J/mol.K$ is referred to as the Universal gas constant

$T=0\°C=273.115K$ represents the absolute standard temperature

$d$ denotes the diameter of helium atoms

$N_{A}=6.0221(10)^{23}/mol$ symbolizes Avogadro's number

$P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}$ indicates absolute standard pressure

<pFrom this, we can solve for $d$ using (1), aiming to determine the radius $r$ of the helium atom:

$d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}$ (2)

$d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}$ (3)

$d=2.0467(10)^{-10}m$ (4)

If the radius equals half of that diameter:

$r=\frac{d}{2}$ (5)

Eventually:

$r=\frac{2.0467(10)^{-10}m}{2}$ (6)

$r=1.0233(10)^{-10}m$ (7)

Nonetheless, we were tasked with finding this radius in nanometers. Knowing $1nm=(10)^{-9}m$ :

$r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm$ (8)

Ultimately:

$r=0.10233nm$ Represents the radius of the helium atom in nanometers.

5 0

2 months ago

What happens to the particles of a liquid when energy is removed from them?

Softa [3030]

Response:

D: The distance among the particles diminishes

Clarification:

Removing energy reduces the activity of molecules, similar to how one slows down in cold temperatures (I believe).

3 0

2 months ago

A new moon is discovered orbiting Neptune with an orbital speed of 9.3 103 m/s. Neptune's mass is 1.0 1026 kg. What is the

Yuliya22 [3333]

The radius of the moon's orbit is calculated as R = 7.715 x 10⁷ m, and the moon's orbital period is T = 14.48 hr. The given orbital speed of the moon is v = 9.3 x 10³ m/s, with Neptune's mass being M = 1.0 x 10²⁶ Kg. The moon's orbital velocity can be expressed using the formula. Therefore, by squaring the equation and resolving for r + h, we calculate: R = GM / v². Upon substituting in, we find R to be 7.715 x 10⁷ m. The relation for the moon's orbital period yields T = 2π/ω and simplistically, T = 2πR/v, where ω = v/r. Following this, we compute T, leading to the conclusion: T = 14.48 hr.

3 0

1 month ago

Two thin lenses with a focal length of magnitude 12.0cm, the first diverging and the second converging, are located 9.00cm apart

kicyunya [3294]

b ) The first lens is a concave lens with a focal length of f₁ = - 12 cm and an object distance of u = - 20 cm. Using the lens formula, 1 / v - 1 / u = 1 / f, we get 1 / v + 1 / 20 = -1 / 12. This leads to 1 / v = - 1 / 20 - 1 / 12, which simplifies to 1 / v = -0.05 - 0.08333, yielding v = -7.5 cm. Consequently, the first image is formed before the first lens, near the object side, which becomes the object for the second lens with a distance of 16.5 cm from the second lens. c ) For the second lens, object distance is u = -16.5 cm, and focal length f₂ = + 12 cm (convex lens). Using the lens formula leads to 1 / v + 1 / 16.5 = 1 / 12, and this results in 1 / v = 1 / 12 - 1 / 16.5, which simplifies to 1 / v = 0.08333 - 0.0606. Finally, we find v = 44 cm (approximately). This image will be formed on the other side of the convex lens, which is 53 cm from the first lens. Magnification by the first lens is v / u = -7.5 / -20 = 0.375. For the second lens, it is v / u = 44 / - 16.5 = -2.67. d ) The total magnification becomes 0.375 x - 2.67 = - 1.00125. The height of the final image is then calculated as 2.50 mm x 1.00125 = 2.503 mm. e ) The final image will be inverted compared to the object since the total magnification is negative.

6 0

29 days ago