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kati45 [8]
3 years ago
9

Whenever you watch a high-flying aircraft overhead, it seems that its sound comes from behind the craft rather than from where y

ou see it. Why is this?
Physics
1 answer:
exis [7]3 years ago
3 0

Answer:

It takes more time for sound to travel to location, thus one "hear" aircraft at  one of the previous positions. Since light travels at much much speed, the plane is seen at its current position.

Explanation:

The speed of the sound in air = 344 m/s

The speed of the light in air = 3×10⁸  m/s

Since the speed of the light is way more than the speed of the sound in air and this when a high flying aircraft is being watched overhead , the person first sees the aircraft because the speed of light is much faster and <u>the sound of the aircraft come from behind the craft rather than from it is being seen because the speed of sound is very less as compared to that of light.</u>

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The intensity of an earthquake wave passing through the earth is measured to be 2.5×106 j/(m2⋅s) at a distance of 43 km from the
vampirchik [111]

r₁ = distance of the point from the source = 43 km = 43000 m

I₁ = intensity of earthquake wave at distance "r₁" = 2.5 x 10⁶ W/m²

r₂ = distance of the point from the source = 1.5 km = 1500 m

I₂ = intensity of earthquake wave at distance "r₂" = ?

we know that , for a constant power , the intensity of wave is inversely proportional to the distance from the source .

I α 1/r²             where I = intensity of wave , r = distance from source

hence we can write

I₁/I₂ = r₂²/r₁²

inserting the values

(2.5 x 10⁶) /I₂ = (1500/43000)²

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Put the steps in the media cycle in the order in which they occur. W. The powers that be communicate to the masses through vario
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One of the most important aspects of civilization is communication.

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2 years ago
A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown down the lane with an initial speed of v = 8.7 m/
luda_lava [24]

Answer:

1)  The magnitude of the angular acceleration = 67.92 rad/s^{2}

2) Magnitude of the linear acceleration = 2.744 m/s^{2}

3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s

4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m

5) the final velocity is 6.21 m/s

Explanation:

the given information :

Bowling mass m = 3.6 kg

Radius = 0.101 m

Initial speed v_{0} = 8.7 m/s

Coefficient of kinetic friction μ = 0.28

1) he magnitude of the angular acceleration of the bowling ball is

F = m a

F_{g}  = μ N  ,   N = m g

F_{g}  = μ m g

1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:

momen inersia of Bowling ball I = (2/5) m R^{2}

torque τ = I α

τ = F R

I α = F R

(2/5) m R^{2}  α = μ m g R

α = (5 μ g / 2R) μ g R

  = (5 x 0.28 x 9.8/ 2 x 0.101)

  = 67.92 rad/s^{2}

2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane

F = - F_{g} , F_{k} is the force of kinetic friction

m a = - μ m g, remove m

the magnitude of linear accelaration is

a = μ g

  = (0.28) (9.8)

  = 2.744 m/s^{2}

3) The bowling ball takes time to begin rolling without slipping:

The linear speed, v_{t} = v_{0} - a t

                            v_{t}  =  v_{0} - μ g t

the angular speed, ω = ω0 + α t

                                ω = ω0 + (5  μ g/2R ) t

v_{t} = ω R

v_{0} - μ g t = ω0 R + (5  μ g/2R ) t R

7 μ g t/2 = v_{0} + ω0 R

hence,

t = (2 v_{0} + ω0 R)/  7 μ g

ω0 = 0 (no initial spin), therefore

t = 2 v_{0} / 7 μ g

 = 2 x 8.7 / 7 (0.28) (9.8)

 = 0.906 s

4) How long it takes for the bowling ball to begin rolling without slipping, S

S = v_{0}  t - (1/2) a t^{2}

  = (8.7) (0.906) - (1/2) (2.744) 0.906^{2}

  = 6.75 m

5) The final velocity

v_{t} = v_{0} - a t

v_{t} = 8.7 - (2.744) (0.906)

v_{t} = 6.21 m/s

4 0
3 years ago
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