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gulaghasi
20 days ago
15

A fire engine is moving south at 35 m/s while blowing its siren at a frequency of 400 Hz.

Physics
1 answer:
ValentinkaMS [3.4K]20 days ago
8 0

To tackle this issue, we will apply the principles linked to the Doppler effect. This effect refers to the alteration in the perceived frequency of any wave when there is relative motion between the source of the waves and the observer. It can be mathematically explained as

f_d= f_s \frac{(v+v_d)}{(v-v_s)}

In this case,

f_d=frequency detected by the receiver

f_s=frequency emitted by the source

v_d=speed of the detector

v_s=speed of the source

v=speed of sound waves

Substituting values yields,

f_d = 400(\frac{(343+18)}{(343-35)})

f_d=422 Hz

Thus, the frequency that passengers would hear is 422Hz

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A 2600-m-high mountain is located on the equator. how much faster does a climber on top of the mountain move than a surfer at a
Maru [3345]

The climber's speed is 0.19 m/s greater than that of the surfer on the beach.

Both individuals are on Earth and share a consistent angular velocity, but their linear speeds differ.

Calculating the seconds in a day gives us t=24*60*60=86400 sec

The linear speed on the beach is computed as

V1=\frac{2πr}{t}

Where t is the duration

Substituting values into the equation leads to

V1=\frac{2π*6.4*10^6}{86400}=465.421 m/s

The mountain's linear speed is determined as

V2=\frac{2π(r+h)}{t}

Inserting values into the equation results in

V2=\frac{2π(6.4*10^6+2600}{86400}=465.61 m/s

Thus, the climber exceeding the surfer's speed is calculated as 465.61-465.421=0.19 m/s

5 0
1 month ago
After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r
Keith_Richards [3271]
Given: t = 103.45 n m. Here’s the explanation: We are provided the refractive index of the cornea as 1.38, and the refractive index of the eye drops as 1.45, with the wavelength for refractive index being 600 nm. Given that the refractive index of eye drops exceeds that of both the cornea and air, the formula applied for constructive interference leads to a minimum thickness of t = 103.45 n m.
6 0
15 days ago
While watching the crane in operation, an observer mentions to you that for a given load there is a maximum angle θmax between 0
ValentinkaMS [3465]
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7 0
17 days ago
Your school science club has devised a special event for homecoming. You've attached a rocket to the rear of a small car that ha
inna [3103]

La force agissant pendant 9 s et la décélération pendant 12 - 9 = 3 s.

Distance totale parcourue = 990 m

vitesse initiale u = 0

Distance parcourue pendant l'accélération

s₁ = 1/2 a 9² où a est l'accélération

= 40.5 a

vitesse finale après 9 s

v = at = 9a

pendant la décélération

v² = u² - 5 x s₂

0 = (9a)² - 5 s₂

s₂ = 16.2 a²

Distance parcourue pendant la décélération = 16.2 a²

s₁ + s₂ = 990

40.5 a + 16.2 a² = 990

16.2 a² + 40.5 a - 990 = 0

a = 6.5

4 0
23 days ago
The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,
Keith_Richards [3271]

Answer:

The initially bent young tree has been straightened by adjusting the tensions of the three guy wires to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Please calculate the force and moment reactions at the trunk's base point O, disregarding the weight of the tree.

C and D are situated 3.1' from the y-axis, while B and C are located 5.4' from the x-axis, and A has a height of 5.2'.

Explanation:

Refer to the attached image.

3 0
1 month ago
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