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Oksi-84
3 months ago
14

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

Physics
1 answer:
inna [3.1K]3 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:

1)

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

2)

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

v=f \lambda

where

f indicates the frequency

\lambda is the wavelength

<pin the="" first="" medium="">

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.

3)

<pwhen a="" wave="" shifts="" from="" one="" medium="" to="" another="" the="" following="" occurs:="">

- 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>
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Early investigators (including Thomas Young) measured the thickness of wool fibers using diffraction. One early instrument used
ValentinkaMS [3465]

Answer:

w= 1.867\times10^{-2}

Explanation:

Provided:

fiber diameter d= 18 μm

screen distance D= 30 cm

wavelength λ= 560 nm

from this, we can determine the fringe width

w=\frac{2\lambda D}{d}

substituting the values yield

w=\frac{2\times560\times10^{-9}\times0.3}{18\times10^{-6}

w= 1.867\times10^{-2}

8 0
3 months ago
A kinesin that is transporting a secretory vesicle uses approximately 80 ATP molecules/s. Each ATP provides a kinesin molecule w
inna [3103]

Answer:

The force is  F = 8*10^{-12} \ N

Explanation:

According to the inquiry, we understand that

     The rate at which ATP molecules are utilized is R = 80 ATP/ s

       The energy yielded by a single ATP molecule is  E_{ATP} = 0.8 * 10^{-19} J

       The kinesin's velocity is  v = 800 nm/s = 800*10^{-9} m/s

The power generated by the ATP in one second can be expressed mathematically as

       P = E_{ATP} * R

After substituting the values

       P = 80 * 0.8*10^{-19 }

       P = 6.4 *10^{-18}J/s

Now this power can be represented mathematically as

       P = F * v

Where  F indicates the force exerted by the kinesin

Therefore

          F = \frac{P}{v}

after substituting input values

            F = \frac{6.4*0^{-18}}{800 *10^{-9}}

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7 0
3 months ago
Two racecars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the
Yuliya22 [3333]

Response:

The acceleration of car 2 is four times that of car 1.

Rationale:

Centripetal acceleration occurs when an object travels in a circular route. It can be expressed as:

a=\dfrac{v^2}{r}

In this scenario, two race cars are moving at consistent speeds around a circular course. Both automobiles are located at an equal distance from the center, but car 2 is operating at twice the speed of car 1.

Thus,

\dfrac{a_1}{a_2}=\dfrac{v_1^2}{v_2^2}

1 and 2 represent the first and second cars, respectively.

v_2=2v_1

Consequently,

\dfrac{a_1}{a_2}=\dfrac{v_1^2}{(2v_1)^2}\\\\\dfrac{a_1}{a_2}=\dfrac{v_1^2}{4v_1^2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{4}\\\\a_2=4\times a_1

Therefore, car 2's acceleration is four times that of car 1.

4 0
3 months ago
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