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2 years ago

A wave which needs a medium (solid, liquid, gas) in order to propagate itself.

Physics

2 answers:

Vadim26 [7]2 years ago

5 0

The correct answer to the question is: Sound wave.

EXPLANATION:

Before going to answer this question, first we have to understand longitudinal wave.

A longitudinal wave is a mechanical wave in which the direction of propagation of wave is parallel to the direction of particle vibration.

This wave propagates through a medium in the form of compression and rarefaction.

Compression are the regions of high pressure zones where the particles of the medium are closely aggregated to each other.

Rarefaction is the region of low pressure where particles of the medium are not so close to each other as compared to compression.

The longitudinal wave can not propagate in space. It always needs a medium for it's propagation. The medium may be solid, liquid or gas .

For instance, sound wave is a longitudinal wave.

Hence, the correct answer of this question is sound wave which needs a medium for it's propagation.

kherson [118]2 years ago

5 0

Answer:

Longitudinal waves

Explanation:

Mechanical waves are those waves which require medium to propagate. It has two types:

i) Transverse waves: These waves depend upon the rigidity of the medium so, they can propagate though solids and liquids but cannot pass through gases.

ii) Longitudinal waves: These waves depend upon the volume elasticity of the medium so, they can propagate through solids, gases, and liquids

Hence, the right answer is longitudinal waves.

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A ball is dropped from an aircraft flying at an altitude of 8,848 meters assuming gravity is 9.8m/s what is the total amount of

Dafna11 [192]

In this question, you're determining the time (t) taken for an object to fall from a distance (d).

The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)

d = 8,848m and g = 9.8m/s

Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)

The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)

Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)

Now for the last step, find the square root of the remaining number:
t = 42.5s

So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.

I hope this helps :)

7 0

2 years ago

We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu

liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

$\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}$

$\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}$

f = 30 cm

using lens formula

$\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})$

$R_1 = R\ and\ R_2 = -R$

$\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})$

$R = (n -1)\ f$

$R = 2(1.5 -1)\ 30$

R = 30 cm

hence, the radii of curvature is 30 cm.

3 0

3 years ago

How far will u go if your wheel' diameter 34in?

Lapatulllka [165]

Technically you can go forever on and on, but maybe your question was like how many rotations in a certain distance?

4 0

2 years ago

A 5.4 g lead bullet moving at 261 m/s strikes a steel plate and stops. If all its kinetic energy is converted to thermal energy

scZoUnD [109]

Answer:

Change in temperature ∆(tita) is 266.097°C

Explanation:

Ok kinectic energy = 1/2MV²

5.4 grams =( 5.4/1000) kilogram

Kinectic energy =( 1/2 )*(5.4/1000)*261²

Kinectic energy = 183.9267 joules

If kinetic energy = thermal energy

183.9267 joules = mc∆(tita)

Where ∆(tita) = change in temperature

And c = 128 J/kg

∆(tita) = 183.9267/((5.4/1000)*128)

∆(tita) = 266.097

∆(tita) = 266.097°C

8 0

2 years ago

How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 4.00

klio [65]

Explanation:

Let us assume that the separation of plate be equal to d and the area of plates is $9 \times 10^{-4} m^{2}$ . As the capacitance of capacitor is given as follows.