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2 years ago

A sound wave traveling eastward through air

1 answer:

4 0

1 - Vibrate east and west

Sound waves are <em>longitudinal waves</em>, that means the vibrations are parallel to the direction of energy transfer. If the energy transfer is going East, then the vibrations will be East and West creating compressions and rarefactions in the air.

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H.e.y w.h.o w.a.nt.s t.o t.a.l.k

butalik [34]

Answer:

nope

Explanation:

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2 years ago

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Inessa05 [86]

The moon lacks an atmosphere compared to the Earth. Hope this helps!

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A long cylindrical capacitor is made of a central wire of radius a = 2.50 mm surrounded by a conducting shell of radius b = 7.50

asambeis [7]

Answer:

The capacitance per unit length is $5.06\times10^{-11}\ F/m$

(b) is correct option.

Explanation:

Given that,

Radius a= 2.50 mm

Radius b=7.50 mm

Dielectric constant = 3.68

Potential difference = 120 V

We need to calculate charge per length for the capacitance

Using formula of charge per length

$\lambda=\dfrac{4\pi\epsilon_{0}\Delta V}{2 ln(\dfrac{r_{2}}{r_{1}})}$

Put the value into the formula

$\lambda=\dfrac{120}{9\times10^{9}\times2 ln(\dfrac{7.50\times10^{-3}}{2.50\times10^{-3}})}$

$\lambda=6.068\times10^{-9}\ C/m$

We know that,

$\lambda=\dfrac{Q}{L}$

We need to calculate the capacitance per unit length

Using formula of capacitance per unit length

$C=\dfrac{\dfrac{Q}{L}}{\Delta V}$

$C=\dfrac{6.068\times10^{-9}}{120}$

$C=5.06\times10^{-11}\ F/m$

Hence, The capacitance per unit length is $5.06\times10^{-11}\ F/m$

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3 years ago

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

Anarel [89]

Answer:

$R = 24.3 m$

Explanation:

As we know that the orbital speed is given as

$v = \sqrt{\frac{GM}{R + h}}$

here we know that

v = 5500 m/s

$R = 4.48 \times 10^6 m$

$h = 630 km$

now we have

$5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}$

$M = 2.34 \times 10^24 kg$

now acceleration due to gravity of planet is given as

$a = \frac{GM}{R^2}$

$a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}$

$a = 7.7 m/s^2$

now range of the projectile on the surface of planet is given as

$R = \frac{v^2 sin2\theta}{g}$

$R = \frac{14.6^2 sin(2\times 30.8)}{7.7}$

$R = 24.3 m$

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3 years ago

What is one way that continental rifts are similar to mid-ocean ridges and what is one that they are different?

adell [148]

Explanation:

Continental rifts and mid-ocean ridges are both features of a divergent plate margin.

In both cases plates are moving away from one another. Therefore they are creating new land masses.

A continental drift like the east African rift valley is where a continent begins to pull apart or diverges.
A mid-ocean ridge is divergent margin in the ocean.

They are different in that, continental rift occurs within the continental plate that are on land.

But:

Mid-ocean ridges are in the oceanic crust in the ocean . They form the largest physiographic structure on the earth surface called the mid-ocean ridge.

learn more:

Descending lithosphere brainly.com/question/9582362

#learnwithBrainly

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2 years ago

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