Physics College

How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 7.1 % increase in the emitted frequency?

Answers

Answer 1

Answer:

v0 = 24.42 m/s (Approx)

Explanation:

Given:

Increase in frequency = 7.1% =

Computation:

Assume n = 100%

n1 = [(v+v0)/(v+v1)]n

[100 + 7.1] = [(344+v0)/(344+0)]100

107.1 = [(344+v0)/(344)]100

v0 = 24.42 m/s (Approx)

Related Questions

two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of 20.g and sphere B has a mass of 10.g.
     If both spheres leave the edge of the table at the same instant, sphere A will land
a.    at some time after sphere B.
b.    at the same time as sphere B.
c.    at some time before sphere B.
d.    There is not enough information to decide.

Answers

Answer:

c. because A will land first becuase its heavier :)

Explanation:

what would the net force be on the box in the problems shown below.( both force and direction). for all four diagrams. please explain answers.

Answers

Answer:
A) object moves 20 N [West] or -20 N [East]
B) object moves 6 N [South] or -6 N [North]
C) object moves 90 N [West] or -90 N [East]
D) object does not move and is at rest*

*Rest means 0

Why:

A)both forces from north and south that are pushing against the object neutralize each other. Assume that north is positive and south is negative: 20 [N] + (-20) [S] = 0
On West and east, you can see that west has a greater force. Assume that west is negative and east is positive: 50 [E] + (-70) [W] = -20 [E]

an unladen swallow flies 5 m/s how far has the bird traveled after 1 minute?

Answers

Answer: 300m

1 minute = 60s

the unladen swallow can fly: 60.5 = 300 (m)

Explanation:

Flying 5 m/s is the same as 300 m/minute. Therefore, it is 300 meters.

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