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Fudgin [204]
3 years ago
5

A trumpeter is placed on a moving sidewalk that moves at a constant velocity of 4.5 m/s. A listener is standing 6.0 m in front o

f the trumpeter on the same sidewalk. The trumpeter plays a note of frequency 290 Hz. The air is still and the speed of sound is What frequency of the note does the listener hear?
Physics
1 answer:
swat323 years ago
7 0
You didn't actually include the speed of sound. But it doesn't matter for this question. If the trumpeter and the listener are on the same moving sidewalk then the distance between them is not changing. The Doppler shift only happens when the distance between the source and the Observer is changing. So the Listener hears the same 290 Hertz that the trumpeter is generating.
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What is the distance between a(-6,3),b(-2,4),c(-8,3)​
Schach [20]

Answer:

Explanation:

Appears to be the vertexes of a triangle.

AB = √(-6 - (-2))² + (3 - 4)²) = √17

AC = √(-6 - (-8))² + (3 - 3)²) = 2

BC = √(-2 - (-8))² + (4 - 3)²) = √37

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nikitadnepr [17]

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Which element has 7 valence electrons?
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3 years ago
When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0
lora16 [44]

Answer:

d=0.165m

Explanation:

Given

m=0.25kg,x_{1}=5cm*\frac{1m}{100cm}=0.05m,x_{2}=4cm*\frac{1m}{100cm}=0.04m,v=2\frac{rev}{s}

The tension of the spring is

F_{k}=K*x_{1}=m*g

K=\frac{m*g}{x_{1}}

K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m

The force in the spring is equal to centripetal force so

F_{c}=\frac{m*v^2}{r}

v=w*r=2\pi*r

But Fc is also

Fc=KxΔr

F_{c}=K*(r-x_{2})

Replacing

m*4\pi^2*r=K*(r-x_{2})

0.25kg*4\pi^2*r=49*(r-0.04m)

r=0.205m

total distance is

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3 0
3 years ago
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