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lozanna [386]
3 years ago
6

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 .
Physics
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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Tourists covered 255 km for a 4-hour ride by car and a 7-hour ride by train. what is the speed of the train, if it is 5 km/h gre
LekaFEV [45]
The distance covered by car is equal to (assuming it is moving by uniform motion) the product between the car's speed and the time of the car ride, 4 h:
S_c = v_c t_c
where
v_c is the car's speed
t_c = 4 h is the duration of the car ride

Similarly, the distance covered by train is equal to the product between the train's speed and the duration of the train ride, 7 h:
S_t = v_t t_t

The total distance covered is S=255 km, which is the sum of the distances covered by car and train:
S=255 km = S_c + S_t
which becomes
255 = 4 v_c + 7 v_t (1)
we also know that the train speed is 5 km/h greater than the car's speed:
v_t = 5 + v_c (2)

If we put (2) into (1), we find
255 = 4v_c + 7(5+v_c)
and if we solve it, we find
v_c = 20 km/h
v_t = 25 km/h

So, the car speed is 20 km/h and the train speed is 25 km/h.

4 0
3 years ago
The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w
Delvig [45]

To develop this problem it is necessary to apply the concept of Frequency based on speed and wavelength.

According to the definition the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity

\lambda = Wavelength

Our value are given by,

v = 345m/s

\lambda = 63cm

Replacing

f = \frac{345}{0.63}

f = 547.61Hz

Therefore the frequency of the tuning fork is 547.61Hz

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2 years ago
A spring with a spring constant value of 2500 StartFraction N over m EndFraction is compressed 32 cm. A 1.5-kg rock is placed on
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Answer:

it's 9m

Explanation:

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2 years ago
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A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

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Answer:

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