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KIM [24]
3 years ago
13

A police car is traveling due east at a speed of 15.0 m/s relative to the earth. You are in a convertible following behind the p

olice car. Your car is also moving due east at 15.0 m/s relative to the earth, so the speed of the police car relative to you is zero. The siren of the police car is emitting sound of frequency 500 Hz. The speed of sound in the still air is 340 m/s.
a.) What is the speed of the sound waves relative to you?
b.) What is the wavelength of the sound waves at your location?
c.) What frequency do you detect?
Physics
1 answer:
Neporo4naja [7]3 years ago
4 0

Answer:

A.) 355 m/s

B.) 0.71 m

C.) 500Hz

Explanation:

Given that a police car is traveling due east at a speed of 15.0 m/s relative to the earth. You are in a convertible following behind the police car. Your car is also moving due east at 15.0 m/s relative to the earth, so the speed of the police car relative to you is zero. The siren of the police car is emitting sound of frequency 500 Hz. The speed of sound in the still air is 340 m/s

a.) What is the speed of the sound waves relative to you?

Since the car is moving away from the observer, the relative velocity will be:

Relative velocity = 340 + 15

Relative velocity = 355 m/s

b.) What is the wavelength of the sound waves at your location?

Using the wave speed formula

V = frequency × wavelength

Make wavelength the subject of formula.

Wavelength = Velocity / frequency

Wavelength = 355/500

Wavelength = 0.71 m

c.) What frequency do you detect?

Fo = Fs ( C + V ) / ( C + v )

Fo = Fs

That is, the frequency of the observer will be equal to the frequency of the source.

Therefore, Fo = 500Hz

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katrin2010 [14]

Answer:

a. 150 g/L

b. 75 g

c. 120 mL

Explanation:

a. 33g/0.22L=150 g/L

b. 33g/0.22L=150 g/L

150 g/L*0.5L=75g

c. 0.22L/33g=0.006667L/g

0.006667L/g*18g=0.12L

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3 years ago
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7 0
3 years ago
A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and
riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0
4 years ago
A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of
PtichkaEL [24]

Answer:

225 N

Explanation:

"Below the horizontal" means he's pushing down at an angle.

Draw a free body diagram of the box.  There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.

Sum of forces in the y direction:

∑F = ma

N − mg − F sin θ = 0

N = F sin θ + mg

Plug in values:

N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)

N = 225 N

8 0
3 years ago
A swimming pool is 4.0 m in depth; a swimmer at this depth feels discomfort in the ear. Calculate the net force on a 0.50-cm-dia
Mashcka [7]

The net force on a 0.50-cm-diameter eardrum is mathematically given as

F= 0.76969 N

<h3>What is the net force on a 0.50-cm-diameter eardrum?</h3>

Generally, the equation for Pressure is  mathematically given as

P = ρgh

Therefore

P= 1000*9.8*4

P= 39200 Pa

Where

A= pi*(0.005/2)^2

Generally, the equation for Net force is  mathematically given as

F = PA

F= 39200 *( pi*(0.005/2)^2)

F= 0.76969 N

In conclusion, The net force is

F= 0.76969 N

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