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3 years ago

How do mechanical waves travel through a medium?

2 answers:

8 0

Particles of the medium move perpendicular to the direction of energy transport. ... Mechanical waves require a medium in order to transport energy. Sound, like any mechanical wave, cannot travel through a vacuum. the particles

EXPLANATION:

uo

max2010maxim [7]3 years ago

7 0

Answer:

sound waves need to travel through a medium such as solids,liquids and gases. hoped this helped

Explanation:

sound waves

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You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of

solong [7]

Answer:

The velocity of the man from the frame of reference of a stationary observer is, V₂ = 5 m/s

Explanation:

Given,

Your velocity, V₁ = 2 m/

The velocity of the person, V₂ =?

The velocity of the person relative to you, V₂₁ = 3 m/s

According to the relative velocity of two

V₂₁ = V₂ -V₁

∴ V₂ = V₂₁ + V₁

On substitution

V₂ = 3 + 2

= 5 m/s

Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s

8 0

3 years ago

Practice

stiv31 [10]

Answer:

The answer is Letter B The car travel at a constant veloc

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3 years ago

Please someone derive 3rd equation of motion u- v= 2sa

zlopas [31]

We know

$\\ \sf\longmapsto a=\dfrac{dv}{dt}$

$\\ \sf\longmapsto a=\dfrac{dv}{dx}.\dfrac{dx}{dt}$

$\\ \sf\longmapsto a=v\dfrac{dv}{dx}$

$\\ \sf\longmapsto adx=vdv$

Integrate

$\\ \sf\longmapsto a{\displaystyle{\int}^x_{x_0}}dx=\displaystyle{\int}_u^v vdv$

$\\ \sf\longmapsto a(x-x_0)=\dfrac{v^2-u^2}{2}$

Here

x-x_0=s

$\\ \sf\longmapsto v^2-u^2=2as$

6 0

3 years ago

Read 2 more answers

An accelerating voltage of 2.42 103 V is applied to an electron gun, producing a beam of electrons originally traveling horizont

Pavlova-9 [17]

Answer:

Part (a) The magnitude of the deflection of electron beam on the screen due to the Earth's gravitational field is 5.97* $10^{-16}$ m.

Part (b) The magnitude of the deflection of electron beam on the screen due to the vertical component of the Earth's magnetic field is $6.19* 10^-3$ m

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3 years ago

The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for

pshichka [43]

Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v ) = 20.5 mi/h = 9.164 M/S

distance ( d ) = 11.6 meter ( m = mass of the car )

Work done = 0.5 m v² = 0.5 * 9.164² * m J = 41.99 m J

Force = ( workdone /distance ) = ( 41.99 m / 11.6 ) = 3.619 m N

Second case

v = 50.2 mi/h = 22.44135 m/s

d = ?

Work done = 0.5 * 22.44² * m J = 251.7768 * m J

Since the braking force remains the same .

3.619 m = ( 251.7768 m / d )

d = 69 .57 meter

7 0

3 years ago