The change in frequency noticed by observers on sub b is : 2.66 Hz before the submarines pass each other.
<u>Given data:</u>
New speed of Sub b ( Vb ) = 12.4 m/s
F₁ = 1416.44 Hz
Frequency of sonar wave ( Fₐ ) = 1400 Hz
Speed of sound in water ( V ) = 1533 m/s
Speed of submarine A ( Va ) = 8.4 m/s
<h3>Determine the change in frequency observed on sub B </h3>
The change in frequency can be calculated using the formula below
F₁' = Fₐ ( V + Vb / V - Vₐ )
= 1400 ( 1533 + 12.4 / 1533 - 8.4 )
= 1400 ( 1545.4 / 1524.6 )
= 1419.1 Hz
Final step : <u>determine the change in frequency </u>
ΔF = F₁' - F₁
= 1419.1 - 1416.44
= 2.66 Hz.
Hence we can conclude that The change in frequency noticed by observers on sub b is : 2.66 Hz.
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Answer:he's not applying force or motion
Explanation:
p=f & m
A satellite orbits the moon for 3685 s. It has an orbital radius of 2008177 m. then the acceleration of the satellite would be 5.832 m/s²
<h3 /><h3>What is a uniform circular motion?</h3>
It is defined as motion when the object is moving in a circle with a constant speed and its velocity is changing with every moment because of the change of direction but the speed of the object is constant in a uniform circular motion
The acceleration during the uniform circular motion is given by the formula
a = v²/r
where a is the acceleration
v is the velocity of the object
r is the radius of the orbit
As given in the problem a satellite orbits the moon for 3685 s. It has an orbital radius of 2008177 m.
velocity of the satellite
v = 2πr/t
v= (2×3.14×2008177)/3685
v = 3422.34 m/s
acceleration of the satellite
a = v²/r
a = 3422.34²/2008177
a = 5.832 m/s²
Thus, the acceleration of the satellite would be 5.832 m/s²
Learn more about uniform circular motion from here
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Answer: Melting, evaporation and sublimation.
