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rosijanka [135]
3 years ago
13

A sound wave has a frequency of 192Hz and travels the length of a football field, 91.4m in 0.267s.  What is the period?

Physics
1 answer:
Vinvika [58]3 years ago
6 0

Answer:1/192 seconds

Explanation:

Period=1÷frequency

Period=1÷192

Period=1/192 seconds

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A child is swinging back and forth with a constant period and amplitude. Somewhere in front of the child, a stationary horn is e
Amanda [17]

Answer:

Explanation:

  We shall apply concept of Doppler's effect of apparent frequency to this problem . Here observer is moving sometimes towards and sometimes away from the source . When observer moves towards the source , apparent frequency is more than real frequency and when the observer moves away from the source , apparent frequency is less than real frequency . The apparent frequency depends upon velocity of observer . The formula for apparent frequency when observer is going away is as follows .

f = f₀ ( V - v₀ ) / V , f is apparent , f₀ is real frequency , V is velocity of sound and v is velocity of observer .

f will be lowest when v₀ is highest .

velocity of observer is highest when he is at the equilibrium position or at middle point .

So apparent frequency is lowest when observer is at the middle point and going away from the source  while swinging to and from before the source of sound .

3 0
3 years ago
The loudness l of a sound, measured in decibels, is given by l=10log10r, where r is the sound's relative intensity. suppose one
marishachu [46]

Answer

given,                              

I is the loudness of sound

I = 10 Log₁₀ r                  

r is relative intensity                    

at when relative intensity is 10⁶        

I = 60 dB                                                  

how much louder when 100 people would be talking together

I = 10 Log₁₀ r                

I = 10 Log₁₀ (10⁶ x 100)  

I = 10 Log₁₀ (10⁸)                

I = 80 dB                      

hence, the intensity will be increased by (80 dB -60 dB) 20 dB when 100 people start talking together.

5 0
3 years ago
Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f
stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2

When s = Db, t = 2t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2

Dividing the two equations

\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db

Hence, Da=(1/4)Db

3 0
3 years ago
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