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shtirl [24]
3 years ago
15

A man is traveling on a bicycle at 14 m/s along a straight road that runs parallel to some railroad tracks. He hears the whistle

of a train that is behind him. The frequency emitted by the train is 840 Hz, but the frequency the man hears is 778 Hz. Take the velocity of sound to be 340 m/s.
a) What frequency is heard by a stationary observer located between the train and the bicycle?

b) What is the speed of the train, and is the train traveling away from or towardthe bicycle?
Physics
1 answer:
deff fn [24]3 years ago
6 0

Answer:

Explanation:

b ) The problem is based on Doppler's effect of sound

f = f₀ x (V - v₀) /( V+v_s)

f is apparent frequency ,f₀ is real frequency , V is velocity of sound , v₀ is velocity of observer going away  , v_s is velocity of source going away

778  = 840 x (340 - 14)/ (340 + v_s)

340 + v_s = 341.18

v_s = 1.18 m /s

it will go  away from   the observer or the cyclist.

speed of train = 1.18 m /s

a )

For a stationary observer v₀ = 0

f = f₀ x V  /( V+v_s)

= 840 x 340 / (340 + 1.180)

= 837 Hz

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Answer:

Explanation:

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The force in a magnetic field is given as

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q(xi + yj + zk) × -1.28k

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28x i×k - 1.28y j×k - 1.28z k×k)

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q( 1.28xj - 1.28y i )

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28y i + 1.28x j)

So comparing comparing coefficients

let compare x axis component

-( 3.70×10−7N )i=-1.28qy i

−3.70×10−7N = -1.28qy

y= -3.7×10^-7/-1.28q

y= -3.7×10^-7/-1.28×-5.90×10^-9)

y=-48.99m/s

y≈-49m/s

Let compare y-axisaxis

7.6×10−7N j = 1.28qx j

7.6×10−7N = 1.28qx

x= 7.6×10^-7/1.28q

x= 7.6×10^-7/1.28×-5.90×10^-9

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a. Then, the velocity of the x component is x= -100.64m/s

b. Also, the velocity component of the y axis is y=-49m/s

c. We will compute

V•F

V=-100.64i -49j

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V•F is

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Answer:

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