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3 years ago

What is the part of the sound wave called where molecules are expanded and are farther apart?

Physics

1 answer:

Lesechka [4]3 years ago

4 0

Answer:

Rarefactions

Explanation:

The sound waves travel in compressions and rarefactions.

As the name suggests compressions are the areas which particles are closer than their usual positioning.

In rarefactions the vibrating particles can be observed further apart than the usual.

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THE LARGEST HORIZONTAL PLATES IS SEPARATED BY 4mm .The plate is at the potencial of -6V . What potencial should be applied to th

Serhud [2]

Answer:

V₂ = -22 V

Explanation:

Electric potential and field are related

ΔV = - E d

where ΔV is the potential difference between the plates, E the electric field and d the separation between the plates

In this exercise we are given the parcionero d = 4 mm = 0.004 m, the potential of one of the plates V1 = -6V and the value of the electric field E = 4000 V / m

V₂- V₁ = - E d

V₂ = - Ed + V₁

V₂ = - 4000 0.004 + (- 6)

V₂ = -16 - 6

V₂ = -22 V

6 0

3 years ago

B) Smartphone, iPad, and tablet are also a kind of computer.

valentinak56 [21]

Answer:

Yes

Explanation:

7 0

3 years ago

the illuminance on a surface is 6 lux and the surface is 4 meters from the light source. What is the intensity of the saurce?

Lelechka [254]

L = illuminance
A = surface
i = intensity

L = i / A ==: i = L * A

i = 6 lux * 4 m^2 = 24 lumen

8 0

3 years ago

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$