When a car approaches you, the sound waves that reach you have a shorter wavelength and a higher frequency. You hear a sound with a higher pitch. When the car moves away from you, the sound waves that reach you have a longer wavelength and lower frequency.
?? ⬇️
An approaching source moves closer during period of the sound wave so the effective wavelength is shortened, giving a higher pitch since the velocity of the wave is unchanged. Similarly the pitch of a receding sound source will be lowered.
The Doppler effect is an effect observed in light and sound waves as they move toward or away from an observer. One simple example of the Doppler effect is the sound of an automobile horn. Picture a person standing on a street corner. A car approaches, blowing its horn.
Comparing two waves of the same wavelength, a higher frequency is associated with faster movement. Comparing two waves of different wavelengths, a higher frequency doesn't always indicate faster movement, although it can. Waves of different wavelengths can have the same frequency.
The pitch of a sound is our ear's response to the frequency of sound. Whereas loudness depends on the energy of the wave. ... The pitch of a sound depends on the frequency while loudness of a sound depends on the amplitude of sound waves.
M)³ / 6 = 4.2e9 m³
<span>so its mass is </span>
<span>M = 3300kg/m³ * 4.2e9m³ = 1.4e13 kg </span>
<span>and so its KE at 16 km/s = 16000 m/s is </span>
<span>KE = ½ * 1.4e13kg * (16000m/s)² = 1.8e21 J
</span># of bombs N = 1.8e21J / 4.0e16J/bomb = 44 234 bombs
<span>give or take.
</span>
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
Answer:
the answer is at the BOTTOM OF THEIR QUESTION
Explanation:
IT IS CORRECT BTW
ANSWER
T₂ = 10.19N
EXPLANATION
Given:
• The mass of the ball, m = 1.8kg
First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,
In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,
The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,
We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,
Solve for T₁,
Now, we use the second equation to find the tension in the horizontal string,
Solve for T₂,
Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.
