Answer:
You have the answer in your comments. I will be copying it so your question doesn't get deleted.
The answers is $0.58
$11.48
the video tape
the new shirt
Answer:it should be Newton’s second law
Explanation: lmk if I’m wrong
To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.
It can be described as

c = Propagation speed of waves in the medium
= Speed of the receiver relative to the medium
= Speed of the source relative to the medium
Frequency emited by the source
The sign depends on whether the receiver or the source approach or move away from each other.
Our values are given by,
Velocity of car
velocity of motor
Velocity of sound
Frequency emited by the source
Replacing we have that



Therefore the frequency that hear the motorcyclist is 601.7Hz
Answer:
To create a second harmonic the rope must vibrate at the frequency of 3 Hz
Explanation:
First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,
f₁ = v/2L
where,
v = speed of wave = 36 m/s
L = Length of rope = 12 m
f₁ = fundamental frequency
Therefore,
f₁ = (36 m/s)/2(12 m)
f₁ = 1.5 Hz
Now the frequency of nth harmonic is given in general, as:
fn = nf₁
where,
fn = frequency of nth harmonic
n = No. of Harmonic = 2
f₁ = fundamental frequency = 1.5 Hz
Therefore,
f₂ = (2)(1.5 Hz)
<u>f₂ = 3 Hz</u>