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3 years ago

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You ride your bike at 12.1 m/s directly away from your neighbor's trumpet sound and toward the sound of another neighbor's tromb

one and find that you hear both instruments at exactly the same pitch. The trumpeter is practicing her middle C at a frequency of 262 Hz . What frequency is the trombonist producing? The speed of sound in air is 337 m/s .

1 answer:

aivan3 [116]3 years ago

8 0

Answer:

243.83 Hz

Explanation:

Given the following :

Recall :

When source is stationary and observe is in motion:

Moving away from the source :

Observed frequency f₀ = f(v-v₀/v)

Moving towards the source :

f₀ = f(v+v₀/v)

Hence ;

v₀ = 12.1 m/s ; f of trumpeter = 262 Hz ; speed of sound (v) = 337m/s

f(v+v₀/v) = 262(v-v₀/v)

f = 262(v-v₀/v) / (v+v₀/v)

f = 262(v-v₀/v) * (v / v+v₀)

f = 262 (v-v₀ / v+v₀)

f = 262 ((337 - 12.1) / (337 + 12.1))

f = 262 (324.9 / 349.1)

f = 262 (0.93067888857)

f = 243.83

f = 244Hz

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d = 105 m

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Speed of a car, v = 21 m/s

We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :

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Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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Answer:

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depth of water when amplitude is down to 1 μV/m

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5 0

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