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3 years ago

Which is not a medium through which a mechanical wave could travel?

Physics

2 answers:

Setler [38]3 years ago

8 0

Mechanical Wave cannot travel in space vaccumes, which would be considered "Empty Space" in your situation.

Tems11 [23]3 years ago

8 0

A mechanical cannot travel through empty space. So option (d) is correct.

A mechanical wave is a wave which needs a material medium for its propagation. For example sound, water waves etc . The medium required by the wave can be a solid, liquid or a gas. Empty space doesn't have any medium, so a mechanical wave cannot travel through empty space.

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A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certa

Aliun [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

The ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

Explanation:

Mathematically Frequency can be represented as

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

Where $\lambda$ is the wavelength and v is the velocity

Now looking at the diagram we see that

For the first frequency we have

Let the wavelength be $\lambda_1 = \lambda$ , and the frequency $F_1 = F$

For the second frequency

Let the wavelength be $\lambda_2 = 2 \lambda$ , and the frequency $F_2 = \frac{F}{2}$

For the third frequency

Let the wavelength be $\lambda_3 = 2\lambda$ , and the frequency $F_3 = \frac{F}{2}$

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

So,

For the first frequency we have

$v_1 = \lambda_1 F_1 = \lambda F$

For the second frequency

$v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F$

For the third frequency

$v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F$

Hence

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

Mathematically angular speed can be represented as

$w = 2 \pi f$

For the first frequency we have

$w_1 = 2\pi F_1 = 2 \pi F$

For the second frequency

$w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2} = \pi F$

For the third frequency

$w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2} = \pi F$

Hence

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

$v = wr$

=> $r = \frac{v}{w}$

Since the linear velocity is constant we have that

$r \ \alpha \ \frac{1}{w}$

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

5 0

3 years ago

A 3,162-kg truck moving with a velocity of 12 m/s to the East hits a 510-kg parked car. The impact causes the car to be set in m

Musya8 [376]

Given:

The mass of the truck is m1 = 3162 kg

The speed of the truck is v1i = 12 m/s in East

The mass of the parked car is m2 = 510 kg

The speed of car is v2i = 0 m/s

The speed of car after collision is v2f = 24 m/s in East

To find the speed of the truck after collision.

Explanation:

The final velocity of the truck will be

$\begin{gathered} m1v1i+m2v2i=m1v1f+m2v2f \\ v1f\text{ =}\frac{m1v1i-m2v2f+m2v2i}{m1} \\ =\frac{3162\times12-510\times24+510\times0}{3162} \\ =8.129\text{ m/s} \end{gathered}$

Thus, the speed of the truck after collision is 8.129 m/s

6 0

1 year ago

What is the amplitude and wavelength of the wave shown below?

masha68 [24]

Answer:

Option B

Explanation:

I did a question like this a long time ago so i'm not sure if its right so sorry if its wrong

3 0

3 years ago

7. An engineer is using a wire that has a resistance of 1.5 . This resistance is too high for the application he is designing. T

Veronika [31]

Answer and Explanation:

We know that resistance $R=\frac{\rho l}{A}$ from the given equation of resistance it is clear that resistance depends on resistivity length and area of the material but we can not change the length because it is given that the length must be 2.5 cm long.